A Unified General Theory of Conic Sections via the Conic Radical

Abstract

In this paper, we bring forth several new general formulae in the classic study of conics in the Analytic Geometry: the coordinates of all vertices and focal points of arbitrary parabolas, ellipses, and hyperbolas; lengths for all latera recta from any non-degenerate conic section; equations describing straight lines whose limited-slope contents stand on exactly equal footing as focal axes, latera recta, and directrices from every non-degenerate conic section; and, respectively, these ones characterizing asymptotes for each non-degenerate hyperbola. All these general results work regardless of whether the conics in question are rotated or not on the Cartesian plane, because all of them depend only on the coefficients of the general conic equation, making the rotation angle irrelevant for the analysis of conic sections.

1. Introduction

Typically, all the traditional textbooks devoted to the study of Analytic Geometry present all the properties and characteristics of the non-rotated conic sections in ${\mathbb{R}}^2$, whereas the general rotated cases are usually overlooked, dedicating just a small part of the exposition to explain how the following standard rotation transformations work:

$$\begin{array}{ccc}\hfill x& =& x^{\prime }cos\theta -y^{\prime }sin\theta ,\hfill \end{array}$$

(1)

$$\begin{array}{ccc}\hfill y& =& x^{\prime }sin\theta +y^{\prime }cos\theta ,\hfill \end{array}$$

(2)

and how to use them to transform the general conic equation:

$$A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0,$$

(3)

with $B\ne 0$, into another equation of the form:

$$A^{\prime }{x\prime }^2+C^{\prime }{y\prime }^2+D^{\prime }{x}^{\prime }+E^{\prime }{y}^{\prime }+F=0,$$

(4)

where $x^{\prime }$ and $y^{\prime }$ are the rotated Cartesian axes generated by some angle $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$, based on the known fact that $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$ whenever $B\ne 0$ in (3) (see [1], pp. 185–188); and after this, these texts stop analyzing the general conic equation and continue with the exposition of other topics on Analytic Geometry, which are not very related to conic sections.

Hence, the main purpose of this paper is to analyze all the non-degenerate loci related to (3), especially when $B\ne 0$, going beyond what a typical Analytic Geometry book usually contains because almost all the results presented here are based on the definition of the “Conic Radical” and the implications of the “General Theorem of Conic Sections” in ${\mathbb{R}}^2$, which allow us to avoid the use of the rotation angle and trigonometric tools to analyze (3) when $B\ne 0$. Indeed, we develop a complete framework for analyzing all non-degenerate conics directly from the coefficients of the general equation, as we provide new general formulae for the following:

• Vertices;
• Foci;
• Latera Recta;
• Directrices;
• Asymptotes;
• Straight lines associated to rotated conics.

The main idea of our paper is to expand the approach presented in [2] so that it is possible to work systematically without requiring the rotation angle, thus introducing the “Conic Radical” and the “General Theorem of Conic Sections”. We intend to cover ellipses, parabolas, and hyperbolas. Concretely, this research represents a comprehensive geometric theory that generalizes classical analytic geometry and extends it with many new formulae.

The structure of this paper is as follows. The next section is a compendium of all the previously known facts that are necessary to achieve the main purpose of this paper; in Section 3, we present some generalities on Analytic Geometry, which are also necessary for the exposition of the results contained in the subsequent sections. Section 4, Section 5 and Section 6 are respectively dedicated to the general analysis of the non-degenerate parabolas, the non-degenerate ellipses, and the non-degenerate hyperbolas. The discussion and conclusions of this research are presented in Section 7.

2. Preliminaries

This section is devoted to reviewing some previously known facts and mathematical tools that are fundamental for the exposition of all the new results presented in this paper.

2.1. Some Well-Known Mathematical Tools

Firstly, consider the matrix of the quadratic equation, which is defined as

$$Q:=\left[\begin{array}{ccc}A& B/2& D/2\\ B/2& C& E/2\\ D/2& E/2& F\end{array}\right],$$

(see p. 30 in [3]) This matrix becomes relevant since (3) can be rewritten as ${\overrightarrow{X}}^{T}Q\overrightarrow{X}=0$, where $\overrightarrow{X}:=(x,y,1)\in {\mathbb{R}}^3$. So, this allows us to define the conic determinant as follows:

$$\Delta :=det\left(Q\right)=\frac{BDE-A{E}^2-C{D}^2-(B^2-4AC)F}{4}.$$

(5)

On the other hand, consider that according to (Ref. [4], pp. 53–54), the inverse transformations of (1) and (2) are given as follows:

$$\begin{array}{cc}\hfill x^{\prime }& =xcos\theta +ysin\theta ,\end{array}$$

(6)

$$\begin{array}{cc}\hfill y^{\prime }& =-xsin\theta +ycos\theta .\end{array}$$

(7)

In addition, the following theorem appears in almost all literature dedicated to Analytic Geometry to calculate the rotation angle $\theta$ of a conic section (see (Ref. [5], p. 68), (Ref. [6], p. 138), (Ref. [7], p. 212) and (Ref. [8], Chapter 1)):

Theorem 1

(First Theorem of the Rotation Angle). Let A, B, C, D, E, $F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a geometric place in ${\mathbb{R}}^2$ different from any single straight line and from any circumference (degenerate or not); if this geometric place is generated by a rotation angle $\theta \in (0,{\displaystyle \frac{\pi }{2}})$, then $B\ne 0$ and also

(i) 

$A=C\iff \theta ={\displaystyle \frac{\pi }{4}}$,

(ii) 

$\theta \in \left(0,{\displaystyle \frac{\pi }{4}}\right)\cup \left({\displaystyle \frac{\pi }{4}},{\displaystyle \frac{\pi }{2}}\right)\iff tan2\theta ={\displaystyle \frac{B}{A-C}}$.

On the other hand, according to (Ref. [6], p. 140), there exist two kinds of conic sections: the ones that have a central point and the ones that do not; the first ones are the circumference, the ellipse, the hyperbola, and all their respective degenerate cases. So, if $\mathcal{C}(h,k)$ is the central point of this kind of conic section, then the two coordinates of this point are respectively given by the following formulae:

$$h=\frac{2CD-BE}{B^2-4AC}\mathrm{and}k=\frac{2AE-BD}{B^2-4AC}.$$

(8)

In contrast, the parabola and its degenerate cases are the only kinds of conic sections that do not have a central point due to $B^2-4AC=0$ in these cases (according to the Discriminant Criterion below), so the formulae in (8) are undefined for these cases. Finally, consider the Sign Function, which is defined for all $a\in \mathbb{R}$ as follows (see (Ref. [9], p. 378)):

$$\mathrm{sgn}\left(a\right):=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}a>0,\hfill \\ 0,\hfill & \mathrm{if}a=0,\hfill \\ -1,\hfill & \mathrm{if}a<0.\hfill \end{array}\right.$$

(9)

Note that this definition immediately implies the following properties of the Sign Function:

• Property 1: $a=\left|a\right|·\mathrm{sgn}\left(a\right)\forall a\in \mathbb{R}$.
• Property 2: $\left|a\right|=a·\mathrm{sgn}\left(a\right)\forall a\in \mathbb{R}$.
• Property 3: ${\mathrm{sgn}}^2\left(a\right)=1\forall a\in \mathbb{R}\setminus \left\{0\right\}$.

2.2. The Definition of the Conic Radical

To prove all the new results that are presented in the subsequent sections of this paper, it is also fundamental to consider the definition of the conic radical of (3), which is given as follows:

$$R:=\eta \sqrt{{(A-C)}^2+B^2},\mathrm{where}\eta =\left\{\begin{array}{cc}\mathrm{sgn}\left(B\right),\hfill & \mathrm{if}B\ne 0,\hfill \\ \mathrm{sgn}(A-C),\hfill & \mathrm{if}B=0.\hfill \end{array}\right.$$

(10)

Note that this definition, (9), and Property 1 imply that R can also be alternatively expressed as

$$R=\left\{\begin{array}{cc}\sqrt{{(A-C)}^2+B^2},\hfill & \mathrm{if}B>0,\hfill \\ A-C,\hfill & \mathrm{if}B=0,\hfill \\ -\sqrt{{(A-C)}^2+B^2},\hfill & \mathrm{if}B<0.\hfill \end{array}\right.$$

In addition, (9) and (10) imply the following facts:

$$\begin{array}{ccc}\hfill R^2& =& {(A-C)}^2+B^2\ge 0\mathrm{in}\mathrm{general},\mathrm{but}R>0\mathrm{when}B\ne 0,\hfill \end{array}$$

(11)

$$\begin{array}{ccc}\hfill \mathrm{sgn}\left(R\right)& =& \mathrm{sgn}\left(B\right)\ne 0,\mathrm{whether}B\ne 0,\hfill \end{array}$$

(12)

$$\begin{array}{ccc}\hfill R& =& 0\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}A-C=0=B,\hfill \end{array}$$

(13)

$$\begin{array}{ccc}\hfill \left|R\right|& =& \sqrt{{(A-C)}^2+B^2}\ge 0,\mathrm{for}\mathrm{all}A,B\mathrm{and}C,\mathrm{which}\mathrm{is}\mathrm{reduced}\mathrm{to}\left|R\right|=|A-C|\ge 0\mathrm{when}B=0.\hfill \end{array}$$

(14)

Likewise, Theorem 1 and (10) imply the following theorem and its subsequent implications (see [2]):

Theorem 2

(Second Theorem of the Rotation Angle). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a rotated locus in ${\mathbb{R}}^2$ whose rotation angle is $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$, and let R its respective conic radical. Then

$$tan\theta =\frac{C-A+R}{B}>0.$$

Hence, if $\theta \in \left[0,{\displaystyle \frac{\pi }{2}}\right)$, then (10) and Theorem 2 imply the following relations:

$$\begin{array}{cc}\hfill sin\theta & =\sqrt{\frac{C-A+R}{2R}}\ge 0,\hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill cos\theta & =\sqrt{\frac{A-C+R}{2R}}>0.\hfill \end{array}$$

(16)

2.3. The General Theorem of Conic Sections and Its Implications

Now consider that if (1) and (2) are substituted in (3), then (10), (13), (15), and (16) imply the following theorem that is fundamental for the study and deep analysis of all conic sections in ${\mathbb{R}}^2$, regardless of whether they are rotated or not, degenerate or not, real or imaginary, etc. So, this subsection is dedicated to reviewing this theorem and all its most relevant implications, which are proven and presented in detail in [2].

Theorem 3

(The General Theorem of Conic Sections). Let A, B, C, D, E, $F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a locus in ${\mathbb{R}}^2$, and let R be its respective conic radical. Then

(i) 

$R=0$ if and only if the locus in question is a single straight line or a circumference (degenerate or not).

(ii) 

$R\ne 0$ if and only if the locus in question is an ellipse (degenerate or not), a parabola (degenerate or not), or a hyperbola (degenerate or not) rotated by an angle $\theta \in [0,{\displaystyle \frac{\pi }{2}})$, whose value is irrelevant to obtain the general equation of this locus without rotations, which is given as follows:

$$\begin{array}{c}\hfill \left(\frac{A+C+R}{2}\right){x^{\prime }}^2+\left(\frac{A+C-R}{2}\right){y^{\prime }}^2+\left[D\sqrt{\frac{1}{2}\left(1+\frac{A-C}{R}\right)}+E\sqrt{\frac{1}{2}\left(1-\frac{A-C}{R}\right)}\right]x^{\prime }\\ \hfill +\left[E\sqrt{\frac{1}{2}\left(1+\frac{A-C}{R}\right)}-D\sqrt{\frac{1}{2}\left(1-\frac{A-C}{R}\right)}\right]y^{\prime }+F=0,\end{array}$$

(17)

where $x^{\prime }$ and $y^{\prime }$ are the rotated Cartesian axes generated by angle θ.

Observe that if $\theta =B=0\ne R$, then this is the only case in which Theorem 3(ii) implies that (3) and (4) are identical, since this is the only case in which $x=x^{\prime }$ and $y=y^{\prime }$. So, this is the reason why from here onward, we say that a locus represented by (3) is “non-rotated” whether $\theta =0$, and is “rotated” whether $\theta \in (0,{\displaystyle \frac{\pi }{2}})$.

Remark 1.

Note that (4) and (17) are identical member to member, which guarantees the following four relations:

$$\begin{array}{c}\hfill A^{\prime }=\frac{A+C+R}{2},C^{\prime }=\frac{A+C-R}{2},D^{\prime }=D\sqrt{\frac{1}{2}\left(1+\frac{A-C}{R}\right)}+E\sqrt{\frac{1}{2}\left(1-\frac{A-C}{R}\right)},E^{\prime }=E\sqrt{\frac{1}{2}\left(1+\frac{A-C}{R}\right)}-D\sqrt{\frac{1}{2}\left(1-\frac{A-C}{R}\right)}.\end{array}$$

Therefore, Remark (1) guarantees that Theorem (3) has the following immediate known implications:

Corollary 1

(Invariants under Rotations). Let A, C, D, E, F, $A^{\prime }$, $C^{\prime }$, $D^{\prime }$, $E^{\prime }\in \mathbb{R}$ be such that (3) and (4) correspond to the same locus on the Cartesian plane, as in Theorem 3(ii). Then, Remark 1 guarantees the following relations:

(i) 

$A^{\prime }+C^{\prime }=A+C$,

(ii) 

${D^{\prime }}^2+{E^{\prime }}^2=D^2+E^2$,

(iii) 

$-4A^{\prime }{C}^{\prime }=B^2-4AC$,

(iv) 

$|A^{\prime }-C^{\prime }|=\sqrt{{(A-C)}^2+B^2}$.

Corollary 2

(The discriminant criterion). Let A, B, C, D, E, $F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a locus in ${\mathbb{R}}^2$, and define the conic discriminant as

$$d:=B^2-4AC.$$

(18)

Then

(i) 

$d<0$ if and only if the locus in question is a circumference, an ellipse or a single point.

(ii) 

$d=0$ if and only if the locus in question is a parabola, a single straight line, or a couple of straight lines with the same slope.

(iii) 

$d>0$ if and only if the locus in question is a hyperbola or a couple of straight lines with different slopes.

On the other hand, Remark 1, Corollary 1, and the definitions given by (10) and (18) also imply the following two lemmas:

Lemma 1.

If $d=0$, Then

(i) 

$B\ne 0$ if and only if $A\ne 0\ne C$,

(ii) 

$\left|R\right|=|A+C|$,

(iii) 

$B\ne 0$ implies $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}(A+C)\ne 0$.

Lemma 2.

If $d<0$, Then

(i) 

$\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}(A+C)\ne 0$,

(ii) 

$\left|R\right|<|A+C|$,

(iii) 

$\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(A^{\prime }\right)=\mathrm{sgn}\left(C^{\prime }\right)=\mathrm{sgn}(A^{\prime }+C^{\prime })\ne 0$.

Finally, some other important implications of Theorem 3 and (10) are the following two results.

Theorem 4

(The Instant Simplification of Rotated and Translated Conic Sections). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a locus on the Cartesian plane that is different from a single straight line or a couple of straight lines with the same slope. Then

(i) 

If $d\ne 0$, then the equation of the locus in question without rotations and translations is given as follows:

$$\left(\frac{A+C+R}{8}\right){x^{″}}^2+\left(\frac{A+C-R}{8}\right){y^{″}}^2=\frac{\Delta }{d},$$

where Δ is as in (5).

(ii) 

If $d=0$ and the locus in question corresponds to a non-degenerate parabola, then $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(R\right)\ne 0$ whenever the equation of this parabola without rotations and translations is given by

$${x^{″}}^2=\frac{1}{A+C}\left(D\sqrt{\frac{C}{A+C}}-E\sqrt{\frac{A}{A+C}}\right)y^{″};$$

so, $\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$ whenever the equation of this parabola without rotations and translations is given by

$${y^{″}}^2=-\frac{1}{A+C}\left(D\sqrt{\frac{C}{A+C}}+E\sqrt{\frac{A}{A+C}}\right)x^{″}.$$

where $x^{″}$ and $y^{″}$ correspond to the Cartesian axes generated by a rotation and a translation at the same time, for some angle $\theta \in \left[0,{\displaystyle \frac{\pi }{2}}\right)$ and some translated origin $O^{\prime }(h,k)\in {\mathbb{R}}^2$.

Remark 2.

(a) Note that Theorem 4(i) guarantees that (3) and (4) correspond to the degenerate cases of circumference, ellipse, and hyperbola whenever $\Delta =0\ne d$; in addition, Theorem 4(ii) implies that (3) and (4) correspond to a non-degenerate parabola whenever $d=0$ and $A{E}^2\ne C{D}^2$.

(b) Remark 1 and Theorem 4(i) also imply that the canonical form of (4), when this equation and (3) correspond to the non-degenerate cases of an ellipse, or a hyperbola, is given by

$${\displaystyle \frac{{x^{″}}^2}{\left(\frac{4\Delta }{A^{\prime }d}\right)}}+{\displaystyle \frac{{y^{″}}^2}{\left(\frac{4\Delta }{C^{\prime }d}\right)}}=1.$$

Theorem 5

(General Formulae for the Eccentricity of Conic Sections). Let A, B, C, D, E, $F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate locus on the Cartesian plane, and let R be its respective conic radical. Then

(i) 

If this locus is a circumference, a parabola, an ellipse, or an equilateral hyperbola, then its eccentricity is $\epsilon =\sqrt{{\displaystyle \frac{2\left|R\right|}{|A+C|+\left|R\right|}}}$.

(ii) 

If this locus is a non-equilateral hyperbola, then its eccentricity is $\epsilon =\sqrt{{\displaystyle \frac{2\left|R\right|}{-\mathrm{sgn}(\Delta )(A+C)+\left|R\right|}}}.$

3. Some Generalities

We devote this section to proving some general facts on conic sections and Analytic Geometry, since most of them are also important for the exposition of the results contained in all three subsequent sections.

3.1. More on the Rotation Angle

Let us begin by exposing some other unknown facts related to the rotation angle. Firstly, the next corollary shows a more concise way to calculate the rotation angle of a conic section than Theorems 1 and 2.

Corollary 3

(Third Theorem of the Rotation Angle). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a rotated locus in ${\mathbb{R}}^2$ whose rotation angle is $\theta \in \left[0,{\displaystyle \frac{\pi }{2}}\right)$, and let R its respective conic radical. Then $tan\theta =\sqrt{{\displaystyle \frac{C-A+R}{A-C+R}}}\ge 0$.

Proof. 

Consider that (10) and (16) guarantee $cos\theta =\sqrt{{\displaystyle \frac{A-C+R}{2R}}}\ne 0$ for all $\theta \in \left[0,{\displaystyle \frac{\pi }{2}}\right)$. Hence, this fact and (15) imply

$$tan\theta ={\displaystyle \frac{sin\theta }{cos\theta }}={\displaystyle \frac{\sqrt{\frac{C-A+R}{2R}}}{\sqrt{\frac{A-C+R}{2R}}}}=\sqrt{\frac{C-A+R}{A-C+R}}\ge 0,\mathrm{for}\mathrm{all}\theta \in \left[0,{\displaystyle \frac{\pi }{2}}\right).$$

This completes the proof. □

Remark 3.

The main difference between Theorem 2 and Corollary 3 is that the latter is slightly more general than the first, because Corollary 3 works for any $\theta \in \left[0,{\displaystyle \frac{\pi }{2}}\right)$, whereas Theorem 2 works only for $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$. In other words, if $\theta \in \left[0,{\displaystyle \frac{\pi }{2}}\right)$, then

$$tan\theta =\left\{\begin{array}{cc}{\displaystyle \frac{C-A+R}{B}}=\sqrt{{\displaystyle \frac{C-A+R}{A-C+R}}}>0\hfill & whenever B\ne 0\ne \theta ,\hfill \\ undefined\hfill & for B=0 by virtue of Theorem \mathsf{2}.\hfill \end{array}\right.$$

In addition, note that $\sqrt{{\displaystyle \frac{C-A+R}{A-C+R}}}\ge 0$ guarantees ${\displaystyle \frac{C-A+R}{A-C+R}}=\left|{\displaystyle \frac{C-A+R}{A-C+R}}\right|\ge 0$. Then Corollary 3 implies $tan\theta =\sqrt{{\displaystyle \frac{C-A+R}{A-C+R}}}=\sqrt{\left|{\displaystyle \frac{C-A+R}{A-C+R}}\right|}={\displaystyle \frac{\sqrt{|C-A+R|}}{\sqrt{|A-C+R|}}}\ge 0$, for all $\theta \in \left[0,{\displaystyle \frac{\pi }{2}}\right)$.

Moreover, note that (15) and (16) guarantee that the four rotation transformations given by (1), (2), (6), and (7) can also be expressed in general without dependence on the rotation angle as follows:

$$\begin{array}{c}\hfill x=x^{\prime }\sqrt{\frac{A-C+R}{2R}}-y^{\prime }\sqrt{\frac{C-A+R}{2R}},y=x^{\prime }\sqrt{\frac{C-A+R}{2R}}+y^{\prime }\sqrt{\frac{A-C+R}{2R}},\\ \hfill x^{\prime }=x\sqrt{\frac{A-C+R}{2R}}+y\sqrt{\frac{C-A+R}{2R}},y^{\prime }=-x\sqrt{\frac{C-A+R}{2R}}+y\sqrt{\frac{A-C+R}{2R}}.\end{array}$$

3.2. A New Alternative to the Discriminant Criterion

Now we present a new criterion that is equivalent to the well-known Discriminant Criterion introduced by Corollary 2. It is based on the definition of the conic radical instead of (18). However, this result is fundamental to establishing its main consequence, Lemma 3 (below). We will extensively use the latter in the sequence.

Proposition 1.

If $d>0$, then $\left|R\right|>|A+C|$.

Proof. 

If $d>0$, then (18) implies $B^2>4AC$; hence,

$$R^2={(A-C)}^2+B^2>A^2-2AC+C^2+4AC=A^2+2AC+C^2={(A+C)}^2.$$

Therefore, $\left|R\right|>|A+C|$. □

The following lemma is immediate from Corollary 2, Lemmas 1(ii) and 2(ii), and Proposition 1.

Lemma 3

(The conic radical criterion). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a locus in ${\mathbb{R}}^2$. Then,

(i) 

$\left|R\right|<|A+C|$ if and only if the locus in question is a circumference, an ellipse, or a single point.

(ii) 

$\left|R\right|=|A+C|$ if and only if the locus in question is a parabola, a single straight line, or a couple of straight lines with the same slope.

(iii) 

$\left|R\right|>|A+C|$ if and only if the locus in question is a hyperbola or a couple of straight lines with different slopes.

3.3. The Equations of the Rotated and Translated Cartesian Axes

We devote this subsection to revealing the equations of the two orthogonal straight lines that correspond to the rotated and translated Cartesian axes $x^{″}$ and $y^{″}$, with respect to the original Cartesian axes x and y (see Figure 1), which are overlooked in the traditional literature of analytic geometry.

Theorem 6.

Let A, B, C, D, E, $F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a locus in ${\mathbb{R}}^2$ translated to some arbitrary point $O^{\prime }(h,k)\in {\mathbb{R}}^2$, with respect to the origin $(0,0)$, and rotated by some angle $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$, with respect to the translated but non-rotated $x^{\prime }$-axis (see Figure 1). So, if $x^{″}$ and $y^{″}$ are the Cartesian axes generated by this translation and this rotation, then the equations of the straight lines that correspond to these axes, with respect to the original Cartesian Axes x and y, are given as follows:

(i) 

$y=\left({\displaystyle \frac{C-A+R}{B}}\right)x+{\displaystyle \frac{Bk-(C-A+R)h}{B}}$, for the $x^{″}$-axis,

(ii) 

$y=\left({\displaystyle \frac{B}{A-C-R}}\right)x+{\displaystyle \frac{(A-C-R)k-Bh}{A-C-R}}$, for the $y^{″}$-axis.

Proof. 

Suppose that the equations of the straight lines that correspond to the axes $x^{″}$ and $y^{″}$ are respectively given by $y=m_{1}x+b_1$ and $y=m_{2}x+b_2$, with $m_1,m_2\in \mathbb{R}\setminus \left\{0\right\}$ due to $B\ne 0\ne \theta$, so, it is clear that $m_2=-{\displaystyle \frac{1}{m_1}}$ since these straight lines are orthogonal ([1], p. 46). In addition, consider that the point $O^{\prime }(h,k)$, which is the translated origin, is also the intersection point of these non-parallel straight lines; thus, its coordinates satisfy both equations at the same time. Whence,

(i)

If the equation $y=m_{1}x+b_1$ corresponds to the $x^{″}$-axis, then it is clear that the rotation angle $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$ of (3) is also the inclination angle of the straight line that corresponds to this axis; thus, Theorem 2 implies $m_1=tan\theta ={\displaystyle \frac{C-A+R}{B}}>0$. On the other hand, if the coordinates of $O^{\prime }$ satisfy the equation in question, then $k=m_{1}h+b_1$. Therefore,

$$b_1=k-m_{1}h=k-\left(\frac{C-A+R}{B}\right)h=\frac{Bk-(C-A+R)h}{B}.$$

(ii)

If the equation $y=m_{2}x+b_2$ corresponds to the $y^{″}$-axis, then Theorem 2 implies $m_2=-{\displaystyle \frac{1}{m_1}}=-{\displaystyle \frac{1}{\left(\frac{C-A+R}{B}\right)}}={\displaystyle \frac{B}{A-C-R}}<0$, thus the inclination angle of the straight line that corresponds to this axis is $\theta +{\displaystyle \frac{\pi }{2}}$ (as in Figure 1). On the other hand, if the coordinates of $O^{\prime }$ also satisfy this other equation, then $k=m_{2}h+b_2$. Therefore,

$$b_2=k-m_{2}h=k-\left(\frac{B}{A-C-R}\right)h=\frac{(A-C-R)k-Bh}{A-C-R}.$$

So, the proof is complete. □

Remark 4.

Note that if Corollary 3 and Remark 3 are applied instead of Theorem 2 in the proof of Theorem 6, and supposing that $\theta \in \left[0,{\displaystyle \frac{\pi }{2}}\right)$ instead of $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$, then, the following equations are obtained:

(i) 

$y=\sqrt{{\displaystyle \frac{C-A+R}{A-C+R}}}x+{\displaystyle \frac{\sqrt{|A-C+R|}k-\sqrt{|C-A+R|}h}{\sqrt{|A-C+R|}}},$ for the $x^{″}$-axis,

(ii) 

$y=-\sqrt{{\displaystyle \frac{A-C+R}{C-A+R}}}x+{\displaystyle \frac{\sqrt{|A-C+R|}k+\sqrt{|C-A+R|}h}{\sqrt{|C-A+R|}}},$ for the $y^{″}$-axis whenever $B\ne 0$, otherwise $x=h$,

which are slightly more general than Theorem 6, because these equations include the non-rotated cases, whereas Theorem 6 works only for the rotated cases due to Remark 3.

Now we present an important result, which can be immediately obtained by substituting (8) and (18) in the equations given in Remark 4.

Theorem 7.

Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a conic section (degenerate or not) in ${\mathbb{R}}^2$ with a central point $\mathcal{C}$, which is located in the translated origin $O^{\prime }(h,k)$, and rotated by some angle $\theta \in \left[0,{\displaystyle \frac{\pi }{2}}\right)$, with respect to the translated but non-rotated $x^{\prime }$-axis (as in Figure 1). So, if $x^{″}$ and $y^{″}$ are the Cartesian axes generated by this translation and this rotation, then the equations of the straight lines that correspond to these axes, with respect to the original Cartesian Axes x and y, are given as follows:

(i) 

$y=\sqrt{{\displaystyle \frac{C-A+R}{A-C+R}}}x+{\displaystyle \frac{\sqrt{|A-C+R|}(2AE-BD)-\sqrt{|C-A+R|}(2CD-BE)}{\sqrt{|A-C+R|}d}},$ for the $x^{″}$-axis,

(ii) 

$y=-\sqrt{{\displaystyle \frac{A-C+R}{C-A+R}}}x+{\displaystyle \frac{\sqrt{|A-C+R|}(2CD-BE)+\sqrt{|C-A+R|}(2AE-BD)}{\sqrt{|C-A+R|}d}},$ for the $y^{″}$-axis whenever $B\ne 0$, otherwise $x=h$.

Remark 5.

Note that Remark 3 also guarantees that the absolute values in all the equations of Remark 4 and Theorem 7 can be omitted without a loss of generality, since $\mathrm{sgn}(A-C+R)=\mathrm{sgn}(C-A+R)\ne 0$. However, it is preferable to use them in general to avoid dealing with unnecessary purely imaginary quantities, which happens whenever $\mathrm{sgn}(A-C+R)=\mathrm{sgn}(C-A+R)=-1$.

The following result sheds light on the location of the axes $x^{″}$ and $y^{″}$ generated by the rotated cases of ellipses and hyperbolas. Note that it can be immediately obtained by substituting (8) and (18) in the equations given by Theorem 6. Additionally, note that all of them are well-defined due to $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$ and Corollary 2 guarantees $B\ne 0\ne d$, whereas (10) guarantees $A-C-R\ne 0$ whenever $B\ne 0$.

Corollary 4.

Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to an ellipse (degenerate or not) or a hyperbola (degenerate or not) in ${\mathbb{R}}^2$, whose central point $\mathcal{C}$ is located in the translated origin $O^{\prime }(h,k)$, and rotated by some angle $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$, with respect to the translated but non-rotated $x^{\prime }$-axis (as in Figure 1). So, if $x^{″}$ and $y^{″}$ are the Cartesian axes generated by this translation and this rotation, then the equations of the straight lines that correspond to these axes, with respect to the original Cartesian Axes x and y, are given as follows:

(i) 

$y=\left({\displaystyle \frac{C-A+R}{B}}\right)x+{\displaystyle \frac{B(2AE-BD)-(C-A+R)(2CD-BE)}{Bd}}$ for the $x^{″}$-axis,

(ii) 

$y=\left({\displaystyle \frac{B}{A-C-R}}\right)x+{\displaystyle \frac{(A-C-R)(2AE-BD)-B(2CD-BE)}{(A-C-R)d}}$, for the $y^{″}$-axis.

The difference between Theorem 7 and Corollary 4 is that the first is more general than the second. Theorem 7 works for all the conic sections with a central point (degenerate or not), regardless of whether they are rotated or not, which also includes the circumference (degenerate or not). Meanwhile, Corollary 4 works only for the rotated cases of ellipses (degenerate or not) and hyperbolas (degenerate or not).

Remark 6.

Observe that Theorem 7 and Corollary 4 do not work for the parabola and its degenerate cases due to Corollary 2(ii), which guarantees $d=0$ in these cases, just as they too lack a central point. However, the vertex of the non-degenerate parabola is taken as the translated origin of this case, whereas the translated origin is undefined for the degenerate cases of the parabola.

Finally, Theorem 6 will be the basis for characterizing the equations of all the straight lines that correspond to the focal axes, the directrices, and the latera recta of any non-degenerate conic section.

3.4. Some Important Facts on Parallel Straight Lines

Additionally, all the results presented in this subsection are also important to characterize the equations of the straight lines that correspond to the directrices and the latera recta of any non-degenerate conic section.

Suppose that $D^2+E^2=0$, which happens if and only if $D=E=0$ due to $D^2\ge 0$ and $E^2\ge 0$ in general. Then, the equation $Dx+Ey+F=0$ is reduced to $F=0$, which cannot represent any single straight line in ${\mathbb{R}}^2$. Thus, we have proved the following result.

Lemma 4.

Let D, E, $F\in \mathbb{R}$ be such that the equation $Dx+Ey+F=0$ corresponds to a single straight line in ${\mathbb{R}}^2$, then $D^2+E^2\ne 0$.

Theorem 8.

Let D, E, $F\in \mathbb{R}$ be such that the equation $Dx+Ey+F=0$ corresponds to a single straight line in ${\mathbb{R}}^2$. Then, the equations that correspond to the two straight lines, which are parallel to the straight line in question and whose distances with respect to this one are $\delta >0$, are given by

$$Dx+Ey+\left(F\pm \delta \sqrt{D^2+E^2}\right)=0.$$

Proof. 

According to (Ref. [10], p. 72), if $Ax+By+C=0$ corresponds to a straight line in ${\mathbb{R}}^2$, and $P(x_1,y_1)$ represents any single point on the Cartesian plane, then the least distance $\delta$ between this point and the straight line in question is given by the following formula:

$$\delta =\left|\frac{A{x}_1+B{y}_1+C}{\sqrt{A^2+B^2}}\right|.$$

(19)

Now suppose that the equation $Dx+Ey+F^{\prime }=0$ corresponds to a straight line ${\mathcal{L}}^{\prime }$, which is parallel to the straight line $\mathcal{L}$ represented by the equation $Dx+Ey+F=0$, and suppose that the distance between both straight lines is $\delta >0$. On the other hand, consider the point $P\left(-{\displaystyle \frac{DF}{D^2+E^2}},-{\displaystyle \frac{EF}{D^2+E^2}}\right)$, whose coordinates are well-defined due to Lemma 4 guarantees $D^2+E^2\ne 0$. Thus, P is a point of the straight line $\mathcal{L}$ since its coordinates satisfy the equation that represents $\mathcal{L}$. Hence, the distance between $\mathcal{L}$ and ${\mathcal{L}}^{\prime }$ is equal to the least distance between P and ${\mathcal{L}}^{\prime }$. Thus, (19) implies

$$\begin{array}{c}\hfill \delta =\left|\frac{D\left(-\frac{DF}{D^2+E^2}\right)+E\left(-\frac{EF}{D^2+E^2}\right)+F^{\prime }}{\sqrt{D^2+E^2}}\right|=\left|\frac{-\frac{(D^2+E^2)F}{D^2+E^2}+F^{\prime }}{\sqrt{D^2+E^2}}\right|=\left|\frac{F^{\prime }-F}{\sqrt{D^2+E^2}}\right|>0,\end{array}$$

which is equivalent to $\pm \delta ={\displaystyle \frac{F^{\prime }-F}{\sqrt{D^2+E^2}}}\ne 0$. Therefore, $F^{\prime }=F\pm \delta \sqrt{D^2+E^2}\ne F$. □

We now give a formula for the degenerate parabola when given its corresponding parallel straight line and its distance to it.

Corollary 5.

Let D, E, $F\in \mathbb{R}$ be such that the equation $Dx+Ey+F=0$ corresponds to a single straight line in ${\mathbb{R}}^2$. Then, the general equation of the degenerate case of the parabola, which corresponds to the couple of straight lines that are parallel to the straight line in question and whose respective distances to this one are $\delta >0$, is given by

$$D^2{x}^2+2DExy+E^2{y}^2+2DFx+2EFy+[F^2-{\delta }^2(D^2+E^2)]=0.$$

Proof. 

According to Theorem 8, the equations of the two straight lines, which are parallel to the straight line $Dx+Ey+F=0$ and whose distances with respect to this one are $\delta >0$, are respectively $Dx+Ey+(F+\delta \sqrt{D^2+E^2})=0$ and $Dx+Ey+(F-\delta \sqrt{D^2+E^2})=0$. So, if (3) is given as follows:

$$\begin{array}{ccc}& & \left[Dx+Ey+(F+\delta \sqrt{D^2+E^2})\right]\left[Dx+Ey+(F-\delta \sqrt{D^2+E^2})\right]\hfill \\ & =& D^2{x}^2+2DExy+E^2{y}^2+2DFx+2EFy+\left[F^2-{\delta }^2(D^2+E^2)\right]=0.\hfill \end{array}$$

Observe that we have that (14) and (18) imply that the absolute value of the conic radical, and the discriminant of this equation are respectively given by $\left|R\right|=\sqrt{{(D^2+E^2)}^2+{\left(2DE\right)}^2}=\sqrt{D^4+2D^2{E}^2+E^4}=\sqrt{{(D^2+E^2)}^2}=|D^2+E^2|\ge 0$, and $d={\left(2DE\right)}^2-4D^2{E}^2=0.$ Finally, note that $D^2{\left(2EF\right)}^2=4D^2{E}^2{F}^2=E^2{\left(2DF\right)}^2,$ as well. Therefore, Corollary 2, Lemma 3, and Remark 2(a) guarantee that (3) corresponds here to a degenerate case of a parabola. □

Now we focus on a particular case of Theorem 8, when $D=m$, $E=-1$, and $F=b$, which cannot correspond to a vertical single straight line due to $E\ne 0$ (see [10], p. 65). Its proof is immediate.

Corollary 6.

Let m, $b\in \mathbb{R}$ be such that the equation $y=mx+b$ corresponds to a non-vertical single straight line in ${\mathbb{R}}^2$. Then, the equations that correspond to the two straight lines, which are parallel to the straight line in question and whose distances with respect to this one are $\delta >0$, are given by $y=mx+\left(b\pm \delta \sqrt{m^2+1}\right).$

Remark 7.

Although Corollary 6 is less general than Theorem 8 because it does not work for the cases of straight lines with an undefined slope, Corollary 6 has an important advantage over Theorem 8. That is, on the Cartesian plane, the straight line represented by the equation $y=mx+\left(b+\delta \sqrt{m^2+1}\right)$ is always located over the straight line whose equation is $y=mx+b$, which is also always located over the straight line whose equation is $y=mx+\left(b-\delta \sqrt{m^2+1}\right)$, since the inequalities $\delta >0$ and $m^2+1\ge 1>0$, for all $m\in \mathbb{R}$, will always guarantee $\delta \sqrt{m^2+1}>0>-\delta \sqrt{m^2+1}$, which implies $mx+\left(b+\delta \sqrt{m^2+1}\right)>mx+b>mx+\left(b-\delta \sqrt{m^2+1}\right)$ for all $m,b,x\in \mathbb{R}$ (see Figure 2, Figure 3 and Figure 4). Meanwhile, although Theorem 8 also works for the straight lines with an undefined slope, the location on the Cartesian plane of the couple of parallel straight lines represented by $Dx+Ey+\left(F\pm \delta \sqrt{D^2+E^2}\right)=0,$ with respect to the straight line that corresponds to the equation $Dx+Ey+F=0$, is ambiguous in general because the locations of these two parallel straight lines depend on the signs of D and E.

Therefore, Corollary 6 is a more useful tool to characterize the straight lines that correspond to the directrices and the latus rectum of all the rotated parabolas than Theorem 8. The next section is a fine example of this fact.

The following is an ancillary result that will help us to characterize straight lines in the sequence.

Proposition 2.

Let $A,B,C\in \mathbb{R}$ be such that $R\ne 0$. Then

(i) 

$m^2+1={\displaystyle \frac{2R}{A-C+R}}$ whether $m=tan\theta$, for any $\theta \in \left[0,{\displaystyle \frac{\pi }{2}}\right)$.

(ii) 

$m^2+1={\displaystyle \frac{2R}{C-A+R}}$ whether $m=-{\displaystyle \frac{1}{tan{\theta }^{\prime }}}$, for any ${\theta }^{\prime }\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$.

Proof. 

(i) If $m=tan\theta$ for any $\theta \in \left[0,{\displaystyle \frac{\pi }{2}}\right)$, then (16) guarantees

$$m^2+1={tan}^2\theta +1\equiv {sec}^2\theta \equiv {\displaystyle \frac{1}{{cos}^2\theta }}={\left(\frac{1}{\sqrt{\frac{A-C+R}{2R}}}\right)}^2={\displaystyle \frac{2R}{A-C+R}},$$

with $A-C+R\ne 0$.

(ii)

If $m=-{\displaystyle \frac{1}{tan\theta }}$ for any $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$, then (15) implies

$$m^2+1={\left(-{\displaystyle \frac{1}{tan\theta }}\right)}^2+1\equiv {cot}^2\theta +1\equiv {csc}^2\theta \equiv {\displaystyle \frac{1}{{sin}^2\theta }}={\left(\frac{1}{\sqrt{\frac{C-A+R}{2R}}}\right)}^2={\displaystyle \frac{2R}{C-A+R}}.$$

Finally, note that $C-A+R\ne 0$, since $\theta \ne 0\ne B$ guarantees $R\ne A-C$ due to (10) and (13).

These items complete the proof. □

4. The Non-Degenerate Parabolas

This section is devoted to studying the properties of the non-degenerate parabola. We start with the length of the latus rectum. We move on to characterize the vertex and the focal axis, and later, we present formulae for its opening direction, its focal point, directrix, and latus rectum. We conclude with several illustrations.

4.1. On the Length of the Latus Rectum of the Parabola

Let us begin by presenting the general formula to calculate the length of the latus rectum of any non-degenerate parabola.

Lemma 5.

Let $A,B,C\in \mathbb{R}$ such that $d=0$, with $A\ne 0$ or $C\ne 0$. Then,

(i) 

${\displaystyle \frac{1}{A+C}}\sqrt{{\displaystyle \frac{A}{A+C}}}={\displaystyle \frac{\sqrt{\left|A\right|}}{\mathrm{sgn}(A+C)\sqrt{{|A+C|}^3}}}$,

(ii) 

${\displaystyle \frac{1}{A+C}}\sqrt{{\displaystyle \frac{C}{A+C}}}={\displaystyle \frac{\sqrt{\left|C\right|}}{\mathrm{sgn}(A+C)\sqrt{{|A+C|}^3}}}$.

Proof. 

Firstly, in order to prove that the equalities

$$\sqrt{\frac{A}{{(A+C)}^3}}=\frac{\sqrt{\left|A\right|}}{\sqrt{{|A+C|}^3}}\mathrm{and}\sqrt{\frac{C}{{(A+C)}^3}}=\frac{\sqrt{\left|C\right|}}{\sqrt{{|A+C|}^3}}$$

hold, consider the following three possible cases:

Case 1. Suppose first that $B\ne 0$. Then Lemma 1(iii) guarantees that $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}(A+C)\ne 0$, whereas the law of signs gives $\mathrm{sgn}(A+C)=\mathrm{sgn}\left({(A+C)}^3\right)$. Therefore, $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left({(A+C)}^3\right)\ne 0$, and we get

$$\frac{A}{{(A+C)}^3}=\left|\frac{A}{{(A+C)}^3}\right|=\frac{\left|A\right|}{{|A+C|}^3}>0\mathrm{and}\frac{C}{{(A+C)}^3}=\left|\frac{C}{{(A+C)}^3}\right|=\frac{\left|C\right|}{{|A+C|}^3}>0,$$

which respectively imply

$$\sqrt{\frac{A}{{(A+C)}^3}}=\frac{\sqrt{\left|A\right|}}{\sqrt{{|A+C|}^3}}\mathrm{and}\sqrt{\frac{C}{{(A+C)}^3}}=\frac{\sqrt{\left|C\right|}}{\sqrt{{|A+C|}^3}}.$$

Case 2. Suppose $A=B=0\ne C$, then

$$\sqrt{\frac{A}{{(A+C)}^3}}=\sqrt{\frac{0}{{(0+C)}^3}}=0=\frac{\sqrt{0}}{\sqrt{{|0+C|}^3}}=\frac{\sqrt{\left|A\right|}}{\sqrt{{|A+C|}^3}}\mathrm{and}$$

$$\sqrt{\frac{C}{{(A+C)}^3}}=\sqrt{\frac{C}{{(0+C)}^3}}=\sqrt{\frac{1}{C^2}}=\sqrt{\frac{1}{{\left|C\right|}^2}}=\sqrt{\frac{\left|C\right|}{{|0+C|}^3}}=\frac{\sqrt{\left|C\right|}}{\sqrt{{|A+C|}^3}}.$$

Case 3. Suppose that $A\ne B=0=C$, then proving $\sqrt{{\displaystyle \frac{A}{{(A+C)}^3}}}={\displaystyle \frac{\sqrt{\left|A\right|}}{\sqrt{{|A+C|}^3}}}$ and $\sqrt{{\displaystyle \frac{C}{{(A+C)}^3}}}={\displaystyle \frac{\sqrt{\left|C\right|}}{\sqrt{{|A+C|}^3}}}$ is similar to case 2.

Finally, note that Property 1 implies ${\displaystyle \frac{1}{A+C}}={\displaystyle \frac{1}{\mathrm{sgn}(A+C)|A+C|}}={\displaystyle \frac{1}{\mathrm{sgn}(A+C)\sqrt{{(A+C)}^2}}}.$ Therefore,

$$\frac{1}{A+C}\sqrt{\frac{A}{A+C}}=\frac{1}{\mathrm{sgn}(A+C)\sqrt{{(A+C)}^2}}\sqrt{\frac{A}{A+C}}=\frac{1}{\mathrm{sgn}(A+C)}\sqrt{\frac{A}{{(A+C)}^3}}=\frac{\sqrt{\left|A\right|}}{\mathrm{sgn}(A+C)\sqrt{{|A+C|}^3}}$$

$$\mathrm{and}\frac{1}{A+C}\sqrt{\frac{C}{A+C}}=\frac{1}{\mathrm{sgn}(A+C)}\sqrt{\frac{C}{{(A+C)}^3}}=\frac{\sqrt{\left|C\right|}}{\mathrm{sgn}(A+C)\sqrt{{|A+C|}^3}}.$$

So, this yields the result. □

Theorem 9

(The general formula for the length of the latus rectum of the parabola). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate parabola; then the length of its latus rectum is given by the following general formula:

$${\ell }_r=\left|\frac{D\sqrt{\left|C\right|}+sE\sqrt{\left|A\right|}}{\sqrt{{|A+C|}^3}}\right|>0,wheres:=\left\{\begin{array}{cc}-1\hfill & if\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(R\right)\ne 0,\hfill \\ 1\hfill & if\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0.\hfill \end{array}\right.$$

Proof. 

If (3) corresponds to a non-degenerate parabola, then it is clear that Theorem 4(ii) and Lemma 5 imply that (3) is equivalent to only one of the following equations:

$${x^{″}}^2=\left(\frac{D\sqrt{\left|C\right|}-E\sqrt{\left|A\right|}}{\mathrm{sgn}(A+C)\sqrt{{|A+C|}^3}}\right)y^{″},\mathrm{whenever}\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(R\right)\ne 0;$$

$$\mathrm{or}{y^{″}}^2=\left(-\frac{D\sqrt{\left|C\right|}+E\sqrt{\left|A\right|}}{\mathrm{sgn}(A+C)\sqrt{{|A+C|}^3}}\right)x^{″},\mathrm{whenever}\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0.$$

This implies that the latus rectum for both cases is respectively given as follows:

$${\ell }_r=\left|\frac{D\sqrt{\left|C\right|}-E\sqrt{\left|A\right|}}{\sqrt{{|A+C|}^3}}\right|>0\mathrm{and}{\ell }_r=\left|\frac{D\sqrt{\left|C\right|}+E\sqrt{\left|A\right|}}{\sqrt{{|A+C|}^3}}\right|>0.$$

Finally, note that these expressions are equivalent to ${\ell }_r=\left|{\displaystyle \frac{D\sqrt{\left|C\right|}+sE\sqrt{\left|A\right|}}{\sqrt{{|A+C|}^3}}}\right|>0$, where $s=-1$ whenever $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(R\right)\ne 0$, and otherwise, $s=1$. □

According to (Ref. [5], pp. 46–47), the least distance between the directrix of a non-degenerate parabola and its vertex, which is also the distance between this one and its focal point, is given in terms of the corresponding latus rectum as $p={\displaystyle \frac{{\ell }_r}{4}}$. Hence, this fact and Theorem 9 immediately yield the following result, which is a general characterization of the minimum distance from the directrix to the vertex of a rotated non-degenerate parabola.

Corollary 7.

Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate parabola. The least distance between its directrix and its vertex, which is also the least distance between this and its focal point, is given by the general formula $p=\left|{\displaystyle \frac{D\sqrt{\left|C\right|}+sE\sqrt{\left|A\right|}}{4\sqrt{{|A+C|}^3}}}\right|,$ where s is defined as in Theorem 9.

4.2. The Vertex and the Focal Axis of the Parabola

According to Remark 6, it is important to know the coordinates of the vertex of a non-degenerate parabola since they are also the coordinates of the translated origin in this case, which cannot be obtained by (8). Hence, these coordinates also become necessary to apply Theorem 6 or Remark 4 to know the equations of the corresponding translated and rotated Cartesian axes, so the following theorem reveals how to obtain these coordinates in general.

Theorem 10

(General Theorem of the Vertex of the Parabola). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate parabola, whose vertex is the point $\mathcal{V}(h,k)\in {\mathbb{R}}^2$. Then, the coordinates of this point are given in general as follows:

(i) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(R\right)\ne 0$, then

$$\begin{array}{c}\hfill h=\frac{\left[A(D^2-2E^2)+4{(A+C)}^{2}F\right]\sqrt{C}-\mathrm{sgn}\left(A\right)\left(2D\sqrt{A^3}+E\sqrt{C^3}\right)E}{4{(A+C)}^2\left(E\sqrt{A}-D\sqrt{C}\right)}andk=\frac{\left[C(2D^2-E^2)-4{(A+C)}^{2}F\right]\sqrt{A}+\mathrm{sgn}\left(A\right)\left(D\sqrt{A^3}+2E\sqrt{C^3}\right)D}{4{(A+C)}^2\left(E\sqrt{A}-D\sqrt{C}\right)}.\end{array}$$

(ii) 

If $\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$, then

$$\begin{array}{c}\hfill h=\frac{\left[A(2E^2-D^2)-4{(A+C)}^{2}F\right]\sqrt{C}+\mathrm{sgn}\left(C\right)\left(E\sqrt{C^3}-2D\sqrt{A^3}\right)E}{4{(A+C)}^2\left(E\sqrt{A}+D\sqrt{C}\right)}andk=\frac{\left[C(2D^2-E^2)-4{(A+C)}^{2}F\right]\sqrt{A}+\mathrm{sgn}\left(C\right)\left(D\sqrt{A^3}-2E\sqrt{C^3}\right)D}{4{(A+C)}^2\left(E\sqrt{A}+D\sqrt{C}\right)}.\end{array}$$

Proof. 

Case 1. Suppose that $B\ne 0$, then Lemma 1(iii) and (12) imply $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}(A+C)\ne 0\ne \mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(R\right)$. So, consider the following two possibilities:

(i)

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}(A+C)=\mathrm{sgn}\left(R\right)\ne 0$, then (17), Remark 1, Corollary 1(i), Theorem 4(ii), and Lemma 3(ii) guarantee that (4) is of the form $A^{\prime }{x^{\prime }}^2+D^{\prime }{x}^{\prime }+E^{\prime }{y}^{\prime }+F=0$ with $A^{\prime }=R=A+C\ne 0=C^{\prime }$, $D^{\prime }=D\sqrt{{\displaystyle \frac{A}{A+C}}}+E\sqrt{{\displaystyle \frac{C}{A+C}}}$ and $E^{\prime }=E\sqrt{{\displaystyle \frac{C}{A+C}}}-D\sqrt{{\displaystyle \frac{A}{A+C}}}\ne 0$. Thus, the canonical form of (4) is ${\left(x^{\prime }+{\displaystyle \frac{D^{\prime }}{2A^{\prime }}}\right)}^2=-{\displaystyle \frac{E^{\prime }}{4A^{\prime }{E}^{\prime }}}\left(y^{\prime }-{\displaystyle \frac{{D^{\prime }}^2-4A^{\prime }F}{4A^{\prime }{E}^{\prime }}}\right)$. Hence, the coordinates of the vertex of the parabola in question on the rotated Cartesian plane are

$$\begin{array}{ccc}\hfill h^{\prime }& =& -\frac{D^{\prime }}{2A^{\prime }}=-\frac{1}{2(A+C)}\left(D\sqrt{\frac{A}{A+C}}+E\sqrt{\frac{C}{A+C}}\right)=-\frac{D\sqrt{A}+E\sqrt{C}}{2(A+C)\sqrt{A+C}},\hfill \\ \hfill k^{\prime }& =& \frac{{D^{\prime }}^2-4A^{\prime }F}{4A^{\prime }{E}^{\prime }}=\frac{{\left(D\sqrt{\frac{A}{A+C}}+E\sqrt{\frac{C}{A+C}}\right)}^2-4(A+C)F}{4(A+C)\left(E\sqrt{\frac{A}{A+C}}-D\sqrt{\frac{C}{A+C}}\right)}=\frac{A{D}^2+2DE\sqrt{AC}+C{E}^2-4{(A+C)}^{2}F}{4\sqrt{{(A+C)}^3}\left(E\sqrt{A}-D\sqrt{C}\right)}.\hfill \end{array}$$

Now consider that (15) and (16) respectively yield $sin\theta =\sqrt{{\displaystyle \frac{C-A+R}{2R}}}=\sqrt{{\displaystyle \frac{C-A+A+C}{2(A+C)}}}=\sqrt{{\displaystyle \frac{C}{A+C}}}$ and $cos\theta =\sqrt{{\displaystyle \frac{A-C+R}{2R}}}=\sqrt{{\displaystyle \frac{A-C+A-C}{2(A+C)}}}=\sqrt{{\displaystyle \frac{A}{A+C}}}.$ Thus, these relations, Property 1 and Lemma 1(iii) imply that the coordinates of the vertex of the parabola in question on the non-rotated Cartesian plane are given as follows:

$$\begin{array}{ccc}\hfill h& =& h^{\prime }cos\theta -k^{\prime }sin\theta =-\frac{D\sqrt{A}+E\sqrt{C}}{2(A+C)\sqrt{A+C}}\sqrt{\frac{A}{A+C}}-\frac{A{D}^2+2DE\sqrt{AC}+C{E}^2-4{(A+C)}^{2}F}{\sqrt{4{(A+C)}^3}\left(E\sqrt{A}-D\sqrt{C}\right)}\sqrt{\frac{C}{A+C}}\hfill \\ & =& \frac{A{D}^2\sqrt{C}-2A{E}^2\sqrt{C}+4{(A+C)}^{2}F\sqrt{C}-2ADE\sqrt{A}-C{E}^2\sqrt{C}}{4{(A+C)}^2\left(E\sqrt{A}-D\sqrt{C}\right)}=\frac{[A(D^2-2E^2)+4{(A+C)}^{2}F]\sqrt{C}-\mathrm{sgn}\left(A\right)\left(2D\sqrt{A^3}+E\sqrt{C^3}\right)E}{4{(A+C)}^2\left(E\sqrt{A}-D\sqrt{C}\right)},\hfill \\ \hfill k& =& h^{\prime }sin\theta +k^{\prime }cos\theta =-\frac{D\sqrt{A}+E\sqrt{C}}{2(A+C)\sqrt{A+C}}\sqrt{\frac{C}{A+C}}+\frac{A{D}^2+2DE\sqrt{AC}+C{E}^2-4{(A+C)}^{2}F}{4\sqrt{{(A+C)}^3}\left(E\sqrt{A}-D\sqrt{C}\right)}\sqrt{\frac{A}{A+C}}\hfill \\ & =& \frac{2C{D}^2\sqrt{A}-C{E}^2\sqrt{A}-4{(A+C)}^{2}F\sqrt{A}+A{D}^2\sqrt{A}+2CDE\sqrt{C}}{4{(A+C)}^2\left(E\sqrt{A}-D\sqrt{C}\right)}=\frac{[C(2D^2-E^2)-4{(A+C)}^{2}F]\sqrt{A}+\mathrm{sgn}\left(A\right)(D\sqrt{A^3}+2E\sqrt{C^3})D}{4{(A+C)}^2\left(E\sqrt{A}-D\sqrt{C}\right)}.\hfill \end{array}$$

(ii)

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}(A+C)=-\mathrm{sgn}\left(R\right)\ne 0$, then (17), Remark 1, Corollary 1(i), Theorem 4(ii), and Lemma 3(ii) guarantee that (4) has the form $C^{\prime }{y^{\prime }}^2+D^{\prime }{x}^{\prime }+E^{\prime }{y}^{\prime }+F=0$, with $A^{\prime }=0\ne C^{\prime }=A+C=-R$, $D^{\prime }=D\sqrt{\frac{C}{A+C}}+E\sqrt{\frac{A}{A+C}}$ and $E^{\prime }=E\sqrt{\frac{C}{A+C}}-D\sqrt{\frac{A}{A+C}}\ne 0.$ Thus, the canonical form of (4) is ${\left(y^{\prime }+\frac{E^{\prime }}{2C^{\prime }}\right)}^2=-\frac{D^{\prime }}{C^{\prime }}\left(y^{\prime }-\frac{{E^{\prime }}^2-4C^{\prime }F}{4C^{\prime }{D}^{\prime }}\right).$ Hence, the coordinates of the vertex of the parabola in question on the rotated Cartesian plane are

$$\begin{array}{ccc}\hfill h^{\prime }& =& \frac{{E^{\prime }}^2-4C^{\prime }F}{4C^{\prime }{D}^{\prime }}=\frac{{\left(E\sqrt{\frac{C}{A+C}}-D\sqrt{\frac{A}{A+C}}\right)}^2-4(A+C)F}{4(A+C)\left(D\sqrt{\frac{C}{A+C}}+E\sqrt{\frac{A}{A+C}}\right)}=\frac{A{D}^2-2DE\sqrt{AC}+C{E}^2-4{(A+C)}^{2}F}{4\sqrt{{(A+C)}^3}\left(E\sqrt{A}+D\sqrt{C}\right)},\hfill \\ \hfill k^{\prime }& =& -\frac{E^{\prime }}{2C}=-\frac{1}{2(A+C)}\left(E\sqrt{\frac{C}{A+C}}-D\sqrt{\frac{A}{A+C}}\right)=\frac{D\sqrt{A}-E\sqrt{C}}{2(A+C)\sqrt{A+C}}.\hfill \end{array}$$

Now consider that (15) and (16) respectively imply $sin\theta =\sqrt{\frac{C-A+R}{2R}}=$$\sqrt{{\displaystyle \frac{C-A+[-(A+C\left)\right]}{2[-(A+C\left)\right]}}}=\sqrt{\frac{A}{A+C}}$ and $cos\theta =\sqrt{\frac{A-C+R}{2R}}=\sqrt{{\displaystyle \frac{A-C+[-(A+C\left)\right]}{2[-(A+C\left)\right]}}}=\sqrt{\frac{C}{A+C}}.$ Thus, these relations and Property 1 imply that the coordinates of the vertex of the parabola in question on the non-rotated Cartesian plane are given as follows:

$$\begin{array}{ccc}\hfill h& =& h^{\prime }cos\theta -k^{\prime }sin\theta =\frac{A{D}^2-2DE\sqrt{AC}+C{E}^2-4{(A+C)}^{2}F}{4\sqrt{{(A+C)}^3}\left(E\sqrt{A}+D\sqrt{C}\right)}\sqrt{\frac{C}{A+C}}-\frac{D\sqrt{A}-E\sqrt{C}}{2(A+C)\sqrt{A+C}}\sqrt{\frac{A}{A+C}}\hfill \\ & =& \frac{2A{E}^2\sqrt{C}-A{D}^2\sqrt{C}-4{(A+C)}^{2}F\sqrt{C}+C{E}^2\sqrt{C}-ADE\sqrt{A}}{4{(A+C)}^2\left(E\sqrt{A}+D\sqrt{C}\right)}=\frac{[A(2E^2-D^2)-4{(A+C)}^{2}F]\sqrt{C}+\mathrm{sgn}\left(C\right)\left(E\sqrt{C^3}-2D\sqrt{A^3}\right)E}{4{(A+C)}^2\left(E\sqrt{A}+D\sqrt{C}\right)},\hfill \\ \hfill k& =& h^{\prime }sin\theta +k^{\prime }cos\theta =\frac{A{D}^2-2DE\sqrt{AC}+C{E}^2-4{(A+C)}^{2}F}{4\sqrt{{(A+C)}^3}\left(E\sqrt{A}+D\sqrt{C}\right)}\sqrt{\frac{A}{A+C}}+\frac{D\sqrt{A}-E\sqrt{C}}{2(A+C)\sqrt{A+C}}\sqrt{\frac{C}{A+C}}\hfill \\ & =& \frac{2C{D}^2\sqrt{A}-C{E}^2\sqrt{A}-4{(A+C)}^{2}F\sqrt{A}+A{D}^2\sqrt{A}-2CDE\sqrt{C}}{4{(A+C)}^2\left(E\sqrt{A}+D\sqrt{C}\right)}=\frac{[C(2D^2-E^2)-4{(A+C)}^{2}F]\sqrt{A}+\mathrm{sgn}\left(C\right)\left(D\sqrt{A^3}-2E\sqrt{C^3}\right)D}{4{(A+C)}^2\left(E\sqrt{A}+D\sqrt{C}\right)}.\hfill \end{array}$$

Case 2. Suppose that $B=0$, then (10) and Theorem 3 imply $R=A-C\ne 0$. Hence, there exist the following two possibilities for (3).

(i)

If $A{x}^2+Dx+Ey+F=0$ with $A\ne C=0\ne E$, then $\mathrm{sgn}\left(A\right)=\mathrm{sgn}(A-C)=\mathrm{sgn}\left(R\right)\ne \mathrm{sgn}\left(C\right)=0$, whereas the canonical form of this equation is ${\left(x+\frac{D}{2A}\right)}^2=-\frac{E}{A}\left(y-\frac{D^2-4AF}{4AE}\right).$ So, the coordinates of the vertex of the parabola in question are $h=-\frac{D}{2A}$ and $k=\frac{D^2-4AF}{4AE}$. Now observe that Property 1 implies

$$\begin{array}{c}\hfill \frac{[A(D^2-2E^2)+4{(A+C)}^{2}F]\sqrt{C}-\mathrm{sgn}\left(A\right)\left(2D\sqrt{A^3}+E\sqrt{C^3}\right)E}{4{(A+C)}^2(E\sqrt{A}-D\sqrt{C})}=\frac{-\mathrm{sgn}\left(A\right)\left(2D\sqrt{A^3}\right)E}{4A^2\left(E\sqrt{A}\right)}=-\frac{D\left(\left|A\right|\mathrm{sgn}\left(A\right)\right)}{2A^2}=-\frac{D}{2A}=h,\\ \hfill \frac{[C(2D^2-E^2)-4{(A+C)}^{2}F]\sqrt{A}+\mathrm{sgn}\left(A\right)\left(D\sqrt{A^3}+2E\sqrt{C^3}\right)D}{4{(A+C)}^2(E\sqrt{A}-D\sqrt{C})}=\frac{-4A^{2}F\sqrt{A}+\mathrm{sgn}\left(A\right)\left(D\sqrt{A^3}\right)D}{4A^2\left(E\sqrt{A}\right)}=\frac{D^2\left(\left|A\right|\mathrm{sgn}\left(A\right)\right)-4A^{2}F}{4A^{2}E}=\frac{D^2-4AF}{4AE}=k.\end{array}$$

(ii)

If $C{y}^2+Dx+Ey+F=0$ with $C\ne A=0\ne D$, then $\mathrm{sgn}\left(C\right)=\mathrm{sgn}(C-A)=\mathrm{sgn}\left(-(A-C)\right)=\mathrm{sgn}(-R)=-\mathrm{sgn}\left(R\right)\ne \mathrm{sgn}\left(A\right)=0$, whereas the canonical form of this equation is ${\left(y+\frac{E}{2C}\right)}^2=-\frac{D}{C}\left(x-\frac{E^2-4CF}{4CD}\right)$. So, the coordinates of the vertex of the parabola in question are $h=\frac{E^2-4CF}{4CD}$ and $k=-\frac{E}{2C}$. Now observe that Property 1 implies

$$\begin{array}{c}\hfill \frac{[A(2E^2-D^2)-4{(A+C)}^{2}F]\sqrt{C}+\mathrm{sgn}\left(C\right)\left(E\sqrt{C^3}-2D\sqrt{A^3}\right)E}{4{(A+C)}^2(E\sqrt{A}+D\sqrt{C})}=\frac{-4C^{2}F\sqrt{C}+\mathrm{sgn}\left(C\right)\left(E\sqrt{C^3}\right)E}{4C^2\left(D\sqrt{C}\right)}=\frac{E^2\left(\left|C\right|\mathrm{sgn}\left(C\right)\right)-4C^{2}F}{4C^{2}D}=\frac{E^2-4CF}{4CD}=h,\\ \hfill \frac{[C(2D^2-E^2)-4{(A+C)}^{2}F]\sqrt{A}+\mathrm{sgn}\left(C\right)\left(D\sqrt{A^3}-2E\sqrt{C^3}\right)D}{4{(A+C)}^2(E\sqrt{A}+D\sqrt{C})}=\frac{\mathrm{sgn}\left(C\right)\left(-2E\sqrt{C^3}\right)D}{4C^2\left(D\sqrt{C}\right)}=-\frac{E\left(\left|C\right|\mathrm{sgn}\left(C\right)\right)}{2C^2}=-\frac{E}{2C}=k.\end{array}$$

Finally, note that all these general formulae for the coordinates of the vertex are well-defined since Theorem 3(ii) and Lemma 3(ii) guarantee $A+C\ne 0$. Meanwhile, Remark 2(a) implies $A{E}^2-C{D}^2=(E\sqrt{A}+D\sqrt{C})(E\sqrt{A}-D\sqrt{C})\ne 0$, which guarantees $E\sqrt{A}+D\sqrt{C}\ne 0\ne E\sqrt{A}-D\sqrt{C}$. □

We now characterize the focal axis of the non-degenerate rotated parabola.

Corollary 8.

Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate parabola. Then, the equation of the straight line that corresponds to its focal axis is given as follows:

(i) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(R\right)\ne 0$, then the equation is $y=-\sqrt{{\displaystyle \frac{A-C+R}{C-A+R}}}x+{\displaystyle \frac{\sqrt{|A-C+R|}h+\sqrt{|C-A+R|}k}{\sqrt{|C-A+R|}}}$, whenever $B\ne 0$; otherwise, $x=h$. Whereas h and k are determined by Theorem 10(i) in general.

(ii) 

If $\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$, then the equation is $y=\sqrt{{\displaystyle \frac{C-A+R}{A-C+R}}}x+{\displaystyle \frac{\sqrt{|A-C+R|}k-\sqrt{|C-A+R|}h}{\sqrt{|A-C+R|}}}$, where h and k are determined by Theorem 10(ii).

Proof. 

(i) If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(R\right)\ne 0$, then Theorem 4(ii) implies that the focal axis of the parabola in question coincides with the $y^{″}$-axis. Hence, Remark 4(ii) determines the equation of the focal axis, where h and k are the coordinates of the vertex, which are given by Theorem 10(i) in this case.

(ii) If $\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$, then Theorem 4(ii) implies that the focal axis of the parabola in question coincides with the $x^{″}$-axis. Hence, Remark 4(i) determines the equation of the focal axis, where h and k are the coordinates of the vertex, which are given by Theorem 10(ii) in this case.

So, the proof is complete. □

The proof of the following result is similar to that of Corollary 8, with the only difference that Theorem 6 is applied instead of Remark 4, whereas (12) implies $\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(R\right)\ne 0$ when $\theta \ne 0\ne B$.

Corollary 9.

Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate parabola rotated by an angle $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$. Then, the equation of the straight line that corresponds to its focal axis is given as follows:

(i) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(B\right)\ne 0$, then the equation is $y=\left({\displaystyle \frac{B}{A-C-R}}\right)x+{\displaystyle \frac{(A-C-R)k-Bh}{A-C-R}}$, where h and k are determined by Theorem 10(i).

(ii) 

If $\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(B\right)\ne 0$, then the equation is $y=\left({\displaystyle \frac{C-A+R}{B}}\right)x+{\displaystyle \frac{Bk-(C-A+R)h}{B}}$, where h and k are determined by Theorem 10(ii).

It is clear that Corollary 8 is more general than Corollary 9 because Corollary 8 works for all the non-degenerate parabolas, rotated or not. Meanwhile, Corollary 9 works only for the rotated parabolas.

4.3. On the Opening Direction of a Rotated Parabola

To know the coordinates of the focal point and the equation of the directrix of any non-degenerate parabola, it is necessary to know first the direction in which the parabola in question opens; for the four cases of the non-rotated parabola, there is well-known criteria to know this (see [10], p. 152), but this is usually overlooked for the rotated parabolas. So, the following lemma shows the criteria to know this for all four possible cases of rotated parabolas.

Lemma 6

(Opening Direction Criteria for Rotated Parabolas). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate parabola, rotated by an angle $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$. Then, the criteria for determining its opening direction are given as follows:

(i) 

It opens right-up if $\mathrm{sgn}\left(A\right)=-\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(E\sqrt{\left|A\right|}+D\sqrt{\left|C\right|}\right)\ne 0$.

(ii) 

It opens left-up if $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(E\sqrt{\left|A\right|}-D\sqrt{\left|C\right|}\right)\ne 0$.

(iii) 

It opens left-down if $\mathrm{sgn}\left(A\right)=-\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(E\sqrt{\left|A\right|}+D\sqrt{\left|C\right|}\right)\ne 0$.

(iv) 

It opens right-down if $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(E\sqrt{\left|A\right|}-D\sqrt{\left|C\right|}\right)\ne 0$.

Proof. 

If (3) corresponds to a non-degenerate parabola rotated by an angle $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$, then $B\ne 0$, so (12) implies $\mathrm{sgn}\left(R\right)=\mathrm{sgn}\left(B\right)\ne 0$, while Lemma 1(iii) guarantees $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}(A+C)\ne 0$. Hence, consider the following two complementary cases:

Case 1. If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}(A+C)=\mathrm{sgn}\left(R\right)=\mathrm{sgn}\left(B\right)\ne 0$, then Theorem 4(ii) and Lemma 5 guarantee that the canonical form of the equation of the parabola in question without rotations and translations is given as ${x^{″}}^2=K_1{y}^{″}$, where

$$\begin{array}{c}\hfill K_1:=-\frac{E\sqrt{\left|A\right|}-D\sqrt{\left|C\right|}}{\mathrm{sgn}\left(A\right)\sqrt{{|A+C|}^3}}\ne 0.\end{array}$$

(20)

So, this parabola opens up or down with respect to the translated and rotated Cartesian Axes $x^{″}$ and $y^{″}$ (see Figure 1), since the $y^{″}$-axis coincides with the focal axis of the parabola in this case. Hence, it opens up whenever $K_1>0$; otherwise, it opens down with respect to these axes. Thus, $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$ implies that the parabola in question opens left-up whenever $K_1>0$; otherwise, it opens right-down with respect to the original Cartesian Axes x and y.

On the other hand, note that the inequality $\sqrt{{|A+C|}^3}>0$, the law of signs, and (20) imply that $K_1>0$ when $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(E\sqrt{\left|A\right|}-D\sqrt{\left|C\right|}\right)\ne 0$, so the parabola opens left-up. Meanwhile, $K_1<0$ when $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(E\sqrt{\left|A\right|}-D\sqrt{\left|C\right|}\right)\ne 0$, so the parabola opens right-down.

Case 2. If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}(A+C)=-\mathrm{sgn}\left(R\right)=-\mathrm{sgn}\left(B\right)\ne 0$, then Theorem 4(ii) and Lemma 5 guarantee that the canonical form of the equation of the parabola in question without rotations and translations is given as ${y^{″}}^2=K_2{x}^{″}$, where

$$\begin{array}{c}\hfill K_2:=-\frac{E\sqrt{\left|A\right|}+D\sqrt{\left|C\right|}}{\mathrm{sgn}\left(A\right)\sqrt{{|A+C|}^3}}\ne 0;\end{array}$$

(21)

so, this parabola opens right or left with respect to the translated and rotated Cartesian Axes $x^{″}$ and $y^{″}$ (see Figure 1), since $x^{″}$-axis coincides with the focal axis of the parabola in this case. Hence, it opens right whenever $K_2>0$; otherwise, it opens left with respect to these axes. Thus, $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$ implies that the parabola in question opens right-up whenever $K_2>0$; otherwise, it opens left-down with respect to the original Cartesian Axes x and y.

On the other hand, note that the inequality $\sqrt{{|A+C|}^3}>0$, the law of signs, and (21) imply that $K_2>0$ when $\mathrm{sgn}\left(A\right)=-\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(E\sqrt{\left|A\right|}+D\sqrt{\left|C\right|}\right)\ne 0$, so the parabola opens right-up. Meanwhile, $K_2<0$ when $\mathrm{sgn}\left(A\right)=-\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(E\sqrt{\left|A\right|}+D\sqrt{\left|C\right|}\right)\ne 0$, so the parabola opens left-down. □

Now that the opening direction of any rotated parabola can be known due to Lemma 6, this information will allow us to obtain more information on rotated parabolas, such as the coordinates of the corresponding focal point and the equation of the corresponding directrix, etc., which are presented in the subsequent subsections, while this information is well-known for the non-rotated cases (Ref. [10], p. 211).

4.4. The Focal Point of the Parabola

Theorem 11

(General Theorem of the Focal Point of the Parabola). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate parabola, and let $\mathcal{V}(h,k)\in {\mathbb{R}}^2$ be the vertex of this parabola, whereas p is the least distance between the vertex and the focal point $\mathcal{F}$ of this parabola. Then, the coordinates of $\mathcal{F}$ are given as follows:

(i) 

If $\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)=-\mathrm{sgn}\left(E\sqrt{\left|A\right|}+D\sqrt{\left|C\right|}\right)\ne 0$, then

$$\mathcal{F}\left(h+p\sqrt{\frac{A-C+R}{2R}},k+p\sqrt{\frac{C-A+R}{2R}}\right).$$

(ii) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(R\right)=-\mathrm{sgn}\left(E\sqrt{\left|A\right|}-D\sqrt{\left|C\right|}\right)\ne 0$, then

$$\mathcal{F}\left(h-p\sqrt{\frac{C-A+R}{2R}},k+p\sqrt{\frac{A-C+R}{2R}}\right).$$

(iii) 

If $\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)=\mathrm{sgn}\left(E\sqrt{\left|A\right|}+D\sqrt{\left|C\right|}\right)\ne 0$, then

$$\mathcal{F}\left(h-p\sqrt{\frac{A-C+R}{2R}},k-p\sqrt{\frac{C-A+R}{2R}}\right).$$

(iv) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(R\right)=\mathrm{sgn}\left(E\sqrt{\left|A\right|}-D\sqrt{\left|C\right|}\right)\ne 0$, then

$$\mathcal{F}\left(h+p\sqrt{\frac{C-A+R}{2R}},k-p\sqrt{\frac{A-C+R}{2R}}\right).$$

Proof. 

(i) If $\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)=-\mathrm{sgn}\left(E\sqrt{\left|A\right|}+D\sqrt{\left|C\right|}\right)\ne 0$, then $\mathrm{sgn}\left(A\right)=-\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(E\sqrt{\left|A\right|}+D\sqrt{\left|C\right|}\right)\ne 0$. For the rotated case, Lemma 6(i) guarantees that the parabola opens right-up on the non-rotated Cartesian plane, so it opens to the right on the rotated Cartesian plane, whereas $\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)=-\mathrm{sgn}\left(D\right)\ne 0=A=B$ for the non-rotated case, so the parabola opens to the right since the canonical form of its equation without translations is ${y^{″}}^2=-{\displaystyle \frac{D}{C}}{x}^{″}$ with $-{\displaystyle \frac{D}{C}}>0$. In any case, suppose that $x^{\prime }$ and $y^{\prime }$ correspond to the Cartesian Axes rotated by $\theta \in \left[0,{\displaystyle \frac{\pi }{2}}\right)$, in which the coordinates of the focal point $\mathcal{F}(x_{\mathcal{F}},y_{\mathcal{F}})$, which is on the non-rotated Cartesian plane, are determined by the point ${\mathcal{F}}^{\prime }(h^{\prime }+p,k^{\prime })$, where $h^{\prime }$ and $k^{\prime }$ are the coordinates of the vertex $\mathcal{V}$ on the rotated Cartesian plane, so (6) and (7) imply $h^{\prime }=hcos\theta +ksin\theta$ and $k^{\prime }=-hsin\theta +kcos\theta$. Hence, these relations, along with (1) and (2) imply

$$\begin{array}{ccc}\hfill x_{\mathcal{F}}& =& (h^{\prime }+p)cos\theta -k^{\prime }sin\theta =(hcos\theta +ksin\theta +p)cos\theta -(-hsin\theta +kcos\theta )sin\theta \hfill \\ & =& h({cos}^2\theta +{sin}^2\theta )+ksin\theta cos\theta -ksin\theta cos\theta +pcos\theta =h+pcos\theta ,\hfill \\ \hfill y_{\mathcal{F}}& =& (h^{\prime }+p)sin\theta +k^{\prime }cos\theta =(hcos\theta +ksin\theta +p)sin\theta +(-hsin\theta +kcos\theta )cos\theta \hfill \\ & =& hsin\theta cos\theta +k({sin}^2\theta +{cos}^2\theta )+psin\theta -hsin\theta cos\theta =k+psin\theta .\hfill \end{array}$$

Finally, from (15) and (16) imply

$$x_{\mathcal{F}}=h+pcos\theta =h+p\sqrt{\frac{A-C+R}{2R}}\mathrm{and}{y}_{\mathcal{F}}=k+psin\theta =k+p\sqrt{\frac{C-A+R}{2R}}.$$

(ii) If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(R\right)=-\mathrm{sgn}\left(E\sqrt{\left|A\right|}-D\sqrt{\left|C\right|}\right)\ne 0$, then $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(E\sqrt{\left|A\right|}-D\sqrt{\left|C\right|}\right)\ne 0$. For the rotated case, Lemma 6(ii) guarantees that the parabola opens left-up on the non-rotated Cartesian plane, so it opens up on the rotated Cartesian plane, whereas $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(R\right)=-\mathrm{sgn}\left(E\right)\ne 0=B=C$ for the non-rotated case, so the parabola opens up since the canonical form of its equation without translations is ${y^{″}}^2=-{\displaystyle \frac{E}{A}}{x}^{″}$ with $-{\displaystyle \frac{E}{A}}>0$. In any case, suppose that $x^{\prime }$ and $y^{\prime }$ correspond to the Cartesian Axes rotated by $\theta \in \left[0,{\displaystyle \frac{\pi }{2}}\right)$, in which the coordinates of the focal point $\mathcal{F}(x_{\mathcal{F}},y_{\mathcal{F}})$, which is on the non-rotated Cartesian plane, are determined by the point ${\mathcal{F}}^{\prime }(h^{\prime },k^{\prime }+p)$, where $h^{\prime }$ and $k^{\prime }$ are the coordinates of the vertex $\mathcal{V}$ on the rotated Cartesian plane, so (6) and (7) imply $h^{\prime }=hcos\theta +ksin\theta ,k^{\prime }=-hsin\theta +kcos\theta$. Hence, these relations, along with (1) and (2) imply

$$\begin{array}{ccc}\hfill x_{\mathcal{F}}& =& h^{\prime }cos\theta -(k^{\prime }+p)sin\theta =(hcos\theta +ksin\theta )cos\theta -(-hsin\theta +kcos\theta +p)sin\theta \hfill \\ & =& h({cos}^2\theta +{sin}^2\theta )+ksin\theta cos\theta -ksin\theta cos\theta -psin\theta =h-psin\theta ,\hfill \\ \hfill y_{\mathcal{F}}& =& h^{\prime }sin\theta +(k^{\prime }+p)cos\theta =(hcos\theta +ksin\theta )sin\theta +(-hsin\theta +kcos\theta +p)cos\theta \hfill \\ & =& hcos\theta sin\theta -hsin\theta cos\theta +k({sin}^2\theta +{cos}^2\theta )+pcos\theta =k+pcos\theta .\hfill \end{array}$$

Finally, (15) and (16) imply

$$x_{\mathcal{F}}=h-psin\theta =h-p\sqrt{\frac{C-A+R}{2R}},\mathrm{and}{y}_{\mathcal{F}}=k+pcos\theta =k+p\sqrt{\frac{A-C+R}{2R}}.$$

(iii) This is similar to the proof of item (i); the only difference is that Lemma 6(iii) for the rotated case and the inequality $-{\displaystyle \frac{D}{C}}<0$ for the non-rotated case guarantee that the parabola in question opens to the left, with respect to the rotated Cartesian Axes, for all $\theta \in \left[0,{\displaystyle \frac{\pi }{2}}\right)$. Therefore, ${\mathcal{F}}^{\prime }(h^{\prime }-p,k^{\prime })$ in this item, which yields the result.

(iv) This is similar to the proof of item (ii); the only difference is that Lemma 6(iv) for the rotated case and the inequality $-{\displaystyle \frac{E}{A}}<0$ for the non-rotated case guarantee that the parabola in question opens down, with respect to the rotated Cartesian Axes, for all $\theta \in \left[0,{\displaystyle \frac{\pi }{2}}\right)$. Therefore, ${\mathcal{F}}^{\prime }(h^{\prime },k^{\prime }-p)$ in this item, which yields the result.

This proves the result. □

4.5. The Equations of the Directrix and the Latus Rectum of a Rotated Parabola

We devote this subsection to presenting how to obtain the equations of a couple of parallel straight lines that respectively correspond to the directrix and the latus rectum of any non-degenerate rotated parabola.

Theorem 12

(Characterization of the Equations of the Directrix and the Latus Rectum of Rotated Parabolas). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate parabola on the Cartesian plane, rotated by an angle $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$, and let $\mathcal{V}(h,k)\in {\mathbb{R}}^2$ be the vertex of this parabola; in addition, consider that the least distance between the corresponding directrix and $\mathcal{V}$, which coincides with the least distance between the vertex and the focal point of this parabola, is given by p. Then, the equations of the straight lines that correspond to the directrix and the latus rectum of the parabola in question are characterized as follows:

(i) 

If it opens right-up, then the equations of the directrix and the latus rectum are respectively given by

$$\begin{array}{c}\hfill y=\left(\frac{B}{A-C-R}\right)x+\left[\frac{(A-C-R)k-Bh}{A-C-R}-p\sqrt{\frac{2R}{R-A+C}}\right]andy=\left(\frac{B}{A-C-R}\right)x+\left[\frac{(A-C-R)k-Bh}{A-C-R}+p\sqrt{\frac{2R}{R-A+C}}\right].\end{array}$$

(ii) 

If it opens left-up, then the equations of the directrix and the latus rectum are respectively given by

$$\begin{array}{ccc}\hfill y=\left(\frac{C-A+R}{B}\right)x+\left[\frac{Bk-(C-A+R)h}{B}-p\sqrt{\frac{2R}{R+A-C}}\right]andy& =& \left(\frac{C-A+R}{B}\right)x+\left[\frac{Bk-(C-A+R)h}{B}+p\sqrt{\frac{2R}{R+A-C}}\right].\hfill \end{array}$$

(iii) 

If it opens left-down, then the equations of the directrix and the latus rectum are respectively given by

$$\begin{array}{c}\hfill y=\left(\frac{B}{A-C-R}\right)x+\left[\frac{(A-C-R)k-Bh}{A-C-R}+p\sqrt{\frac{2R}{R-A+C}}\right]andy=\left(\frac{B}{A-C-R}\right)x+\left[\frac{(A-C-R)k-Bh}{A-C-R}-p\sqrt{\frac{2R}{R-A+C}}\right].\end{array}$$

(iv) 

If it opens right-down, then the equations of the directrix and the latus rectum are respectively given by

$$\begin{array}{c}\hfill y=\left(\frac{C-A+R}{B}\right)x+\left[\frac{Bk-(C-A+R)h}{B}+p\sqrt{\frac{2R}{R+A-C}}\right]andy=\left(\frac{C-A+R}{B}\right)x+\left[\frac{Bk-(C-A+R)h}{B}-p\sqrt{\frac{2R}{R+A-C}}\right].\end{array}$$

Proof. 

Suppose that the equation $y=mx+b$ corresponds to the tangent line of the parabola in question, whose vertex is the corresponding point of tangency. Thus, it is clear that this straight line is perpendicular to the focal axes of the parabola, whereas it is parallel to the two straight lines that respectively correspond to the directrix and the latus rectum of the parabola, and p is the distance between the tangent line and each of these two parallel straight lines. Hence, consider the following four possible cases:

(i)

If the parabola opens right-up, then the tangent line coincides with the $y^{″}$-axis (see Figure 1). Hence, Lemma 6(i), Corollary 9(ii) and Theorem 6(ii) imply $m=-{\displaystyle \frac{1}{tan\theta }}={\displaystyle \frac{B}{A-C-R}}$ and $b={\displaystyle \frac{(A-C-R)k-Bh}{A-C-R}}.$ In addition, Proposition 2(ii) guarantees that $\sqrt{m^2+1}=\sqrt{{\displaystyle \frac{2R}{C-A+R}}}$. Therefore, the opening direction, Corollary 6, and Remark 7 guarantee that $y=mx+\left(b+p\sqrt{m^2+1}\right)$ corresponds to the latus rectum, whereas $y=mx+\left(b-p\sqrt{m^2+1}\right)$ corresponds to the directrix.

(ii)

If the parabola opens left-up, then the tangent line coincides with the $x^{″}$-axis (see Figure 1). Hence, Lemma 6(ii), Corollary 9(i), and Theorem 6(i) imply $m={\displaystyle \frac{C-A+R}{B}}$ and $b={\displaystyle \frac{Bk-(C-A+R)h}{B}}$. In this case, Proposition 2(i) guarantees $\sqrt{m^2+1}=\sqrt{{\displaystyle \frac{2R}{A-C+R}}}.$ Therefore, the opening direction, Corollary 6 and Remark 7 guarantee that $y=mx+\left(b+p\sqrt{m^2+1}\right)$ corresponds to the latus rectum, whereas $y=mx+\left(b-p\sqrt{m^2+1}\right)$ corresponds to the directrix.

(iii)

This is similar to the proof of item (i); the only difference is that the equations of the directrix and the latus rectum are interchanged due to the opening direction.

(iv)

This is similar to the proof of item (ii); the only difference is that the equations of the directrix and the latus rectum are interchanged, as well, due to the opening direction.

This completes the proof. □

4.6. Examples of Rotated Parabolas

Example 1.

Suppose that $225x^2-240xy+64y^2+986x-1836y-3179=0$. Then $R=\mathrm{sgn}(-240)\sqrt{{(225-64)}^2+{(-240)}^2}=-289$ and $|A+C|=|225+64|=289=\left|R\right|$. In addition, $A{E}^2=\left(225\right){(-1836)}^2=\mathrm{758,451,600}\ne C{D}^2=\left(64\right){\left(986\right)}^2=\mathrm{62,220,544}$. Hence, Lemma 3(ii) and Remark 2(a) guarantee that the given equation corresponds to a non-degenerate parabola. Meanwhile, $E\sqrt{\left|A\right|}+D\sqrt{\left|C\right|}=-1836\sqrt{\left|225\right|}+986\sqrt{\left|64\right|}=-\mathrm{19,652}$. So, $\mathrm{sgn}\left(A\right)=-\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(E\sqrt{\left|A\right|}+D\sqrt{\left|C\right|}\right)=1$, thus Lemma 6(i) implies that this parabola opens right-up. On the other hand, Theorem 9 implies $s=1$ due to $\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)=1$, so ${\ell }_r=\left|{\displaystyle \frac{-1836\sqrt{\left|64\right|}+\left(1\right)\left(986\right)\sqrt{\left|225\right|}}{\sqrt{{|225+64|}^3}}}\right|=4$, and Corollary 7 implies $p={\displaystyle \frac{{\ell }_r}{4}}={\displaystyle \frac{4}{4}}=1$. Additionally, Theorem 10(ii) implies that the coordinates of the vertex of the parabola are given by

$$\begin{array}{ccc}\hfill h& =& \frac{[225\left(2{(-1836)}^2-{986}^2{)-4(225+64)}^2(-3179)\right]\sqrt{64}+\mathrm{sgn}\left(64\right)\left[-1836\sqrt{{64}^3}-2\left(986\right)\sqrt{{225}^3}\right](-1836)}{4{(225+64)}^2\left(-1836\sqrt{225}+986\sqrt{64}\right)}=-5,\hfill \\ \hfill k& =& \frac{\left[64\left(2{\left(986\right)}^2-{(-1836)}^2\right)-4{(225+64)}^2(-3179)\right]\sqrt{225}+\mathrm{sgn}\left(\sqrt{64}\right)\left[986\sqrt{{225}^3}-2(-1836)\sqrt{{64}^3}\right]\left(986\right)}{4{(225+64)}^2\left(-1836\sqrt{225}+986\sqrt{64}\right)}=-3.\hfill \end{array}$$

So $\mathcal{V}(-5,-3)$, whereas Corollary 9(ii) implies that the equation of the focal axis is

$$y=\left[\frac{64-225+(-289)}{(-240)}\right]x+\frac{-240(-3)-[64-225+(-289\left)\right](-5)}{(-240)},$$

which is reduced to $y={\displaystyle \frac{15}{8}}x+{\displaystyle \frac{51}{8}}$. Now consider that

$$\begin{array}{c}\hfill \sqrt{\frac{C+A+R}{2R}}=\sqrt{\frac{64+225+(-289)}{2(-289)}}=\frac{15}{17}and\sqrt{\frac{A-C+R}{2R}}=\sqrt{\frac{225-64+(-289)}{2(-289)}}=\frac{8}{17}.\end{array}$$

Thus, Theorem 11(i) implies that the coordinates of the focal point of the parabola are

$$x_{\mathcal{F}}=-5+\left(1\right)\left(\frac{8}{17}\right)=-\frac{77}{17}and{y}_{\mathcal{F}}=-3+\left(1\right)\left(\frac{15}{17}\right)=-\frac{36}{17},so\mathcal{F}\left(-\frac{77}{17},-\frac{36}{17}\right).$$

Likewise, Theorem 12(i) implies that the equations of the directrix and the latus rectum are respectively given as follows:

$$\begin{array}{ccc}\hfill y& =& \left[\frac{(-240)}{225-64-(-289)}\right]x+\left[\frac{\left(225-64-(-289)\right)(-3)-(-240)(-5)}{225-64-(-289)}-\left(1\right)\sqrt{\frac{2(-289)}{-289-225+64}}\right],\hfill \\ \hfill y& =& \left[\frac{(-240)}{225-64-(-289)}\right]x+\left[\frac{\left(225-64-(-289)\right)(-3)-(-240)(-5)}{225-64-(-289)}+\left(1\right)\sqrt{\frac{2(-289)}{-289-225+64}}\right],\hfill \end{array}$$

which are reduced to $y=-{\displaystyle \frac{8}{15}}x-{\displaystyle \frac{136}{15}}$ and $y=-{\displaystyle \frac{8}{15}}x-{\displaystyle \frac{34}{15}}.$ Finally, note that the rotation angle was completely unneeded to obtain all these results. However, if it is required, then it can be obtained by applying Corollary 3 as follows: $tan\theta =\sqrt{{\displaystyle \frac{64-225+(-289)}{225-64+(-289)}}}={\displaystyle \frac{15}{8}}$, thus $\theta =arctan\left({\displaystyle \frac{15}{8}}\right)\approx 61.93°$. See Figure 5.

Example 2.

Suppose that $-x^2-2xy-y^2+2x-6y+7=0$. Then

$$R=\mathrm{sgn}(-2)\sqrt{{[-1-(-1)]}^2+{(-2)}^2}=-2$$

and $|A+C|=|-1+(-1\left)\right|=2=\left|R\right|$. In addition, $A{E}^2=(-1){(-6)}^2=-36\ne C{D}^2=(-1){\left(2\right)}^2=-4$. Hence, Lemma 3(ii) and Remark 2(a) guarantee that the given equation corresponds to a non-degenerate parabola. Meanwhile, $E\sqrt{\left|A\right|}-D\sqrt{\left|C\right|}=-6\sqrt{|-1|}-2\sqrt{|-1|}=-8$. So, $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(E\sqrt{\left|A\right|}-D\sqrt{\left|C\right|}\right)=-1$. Thus Lemma 6(iv) guarantees that this parabola opens right-down. On the other hand, Theorem 9 implies $s=-1$ due to $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(R\right)=-1$. So ${\ell }_r=\left|{\displaystyle \frac{-6\sqrt{|-1|}+(-1)\left(2\right)\sqrt{|-1|}}{\sqrt{{|-1+(-1)|}^3}}}\right|=2\sqrt{2}$, and Corollary 7 implies $p={\displaystyle \frac{{\ell }_r}{4}}={\displaystyle \frac{2\sqrt{2}}{4}}={\displaystyle \frac{\sqrt{2}}{2}}$. Additionally, Theorem 10(i) implies that the coordinates of the vertex of the parabola are given by

$$\begin{array}{ccc}\hfill h& =& \frac{\left[(-1)\left(2^2-2{(-6)}^2\right)+4{\left(-1+(-1)\right)}^2\left(7\right)\right]\sqrt{-1}-\mathrm{sgn}(-1)\left[2\left(2\right)\sqrt{{(-1)}^3}+(-6)\sqrt{{(-1)}^3}\right](-6)}{4{[-1+(-1)]}^2[-6\sqrt{-1}-2\sqrt{-1}]}=\frac{180i+12i}{(-128i)}=-\frac{3}{2},\hfill \\ \hfill k& =& \frac{\left[(-1)\left(2{\left(2\right)}^2-{(-6)}^2\right)-4{\left(-1+(-1)\right)}^2\left(7\right)\right]\sqrt{-1}+\mathrm{sgn}(-1)\left[\left(2\right)\sqrt{{(-1)}^3}+2(-6)\sqrt{{(-1)}^3}\right]\left(2\right)}{4{[-1+(-1)]}^2[-6\sqrt{-1}-2\sqrt{-1}]}=\frac{-84i+20i}{(-128i)}=\frac{1}{2}.\hfill \end{array}$$

So $\mathcal{V}\left(-{\displaystyle \frac{3}{2}},{\displaystyle \frac{1}{2}}\right)$, whereas Corollary 9(i) implies that the equation of the focal axis is

$$y=\left[\frac{(-2)}{-1-(-1)-(-2)}\right]x+\frac{[-1-(-1)-(-2\left)\right](1/2)-(-2)(-3/2)}{-1-(-1)-(-2)}.$$

which is reduced to $y=-x-1$. Now consider that $\sqrt{{\displaystyle \frac{A-C+R}{2R}}}=\sqrt{{\displaystyle \frac{C-A+R}{2R}}}=\sqrt{{\displaystyle \frac{-1-(-1)+(-2)}{2(-2)}}}={\displaystyle \frac{\sqrt{2}}{2}}$, thus Theorem 11(iv) implies that the coordinates of the focal point of the parabola are given by

$$\begin{array}{c}\hfill x_{\mathcal{F}}=-\frac{3}{2}+\frac{\sqrt{2}}{2}\left(\frac{\sqrt{2}}{2}\right)=-1and{y}_{\mathcal{F}}=\frac{1}{2}-\frac{\sqrt{2}}{2}\left(\frac{\sqrt{2}}{2}\right)=0,so\mathcal{F}(-1,0).\end{array}$$

Likewise, Theorem 12(iv) implies that the equations of the directrix and the latus rectum are respectively given as follows:

$$\begin{array}{c}\hfill y=\left[\frac{-1-(-1)+(-2)}{(-2)}\right]x+\left[\frac{-2(1/2)-\left(-1-(-1)+(-2)\right)(-3/2)}{(-2)}+\frac{\sqrt{2}}{2}\sqrt{\frac{2(-2)}{-1-(-1)+(-2)}}\right],\\ \hfill y=\left[\frac{-1-(-1)+(-2)}{(-2)}\right]x+\left[\frac{-2(1/2)-\left(-1-(-1)+(-2)\right)(-3/2)}{(-2)}-\frac{\sqrt{2}}{2}\sqrt{\frac{2(-2)}{-1-(-1)+(-2)}}\right],\end{array}$$

which are reduced to $y=x+3$ and $y=x+1$. Finally, note that the rotation angle was completely unneeded to obtain all these results. However, if it is required, then it can be obtained by applying Corollary 3 as follows: $tan\theta =\sqrt{{\displaystyle \frac{-1-(-1)+(-2)}{-1-(-1)+(-2)}}}=1$, thus $\theta =arctan\left(1\right)=45°.$ See Figure 6.

5. The Non-Degenerate Ellipses

While there are essentially four different cases of non-degenerate parabolas depending on the opening direction, there are only two possible cases of ellipses, depending on which of the Cartesian Axes $x^{″}$ or $y^{″}$ is the corresponding focal axis (see Figure 1). This section is devoted to presenting the two cases of the non-degenerate ellipse. We begin by studying the length of the semiaxes and the latera recta, then we introduce characterizations of the equation of the focal axis, the extreme and focal points, the directrices, and the latera recta. We conclude the section with several illustrations.

5.1. On the Lengths of the Semiaxes and the Latera Recta of the Ellipse

We start we an obvious, but important observation that will be extensively used in the sequel.

Remark 8.

Note that $d<0$ implies $-d=\left|d\right|=\sqrt{d^2}$.

The following is an important ancillary result concerning the lengths of the semi-major and semi-minor axes of the non-degenerate ellipse.

Lemma 7.

Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate ellipse on the Cartesian plane, so if a and b are, respectively, the lengths of its semi-major and semi-minor axes, and c is the least distance between the central point of the ellipse and each of its two focal points. Then

(i) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(R\right)\ne 0$, then $a=2\sqrt{{\displaystyle \frac{2\Delta }{(A+C-R)d}}}>b=2\sqrt{{\displaystyle \frac{2\Delta }{(A+C+R)d}}}>0$ and $c=-{\displaystyle \frac{4\sqrt{-R\Delta }}{d}}>0$.

(ii) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$, then $a=2\sqrt{{\displaystyle \frac{2\Delta }{(A+C+R)d}}}>b=2\sqrt{{\displaystyle \frac{2\Delta }{(A+C-R)d}}}>0$ and $c=-{\displaystyle \frac{4\sqrt{R\Delta }}{d}}>0$.

Proof. 

According to [2], (3) corresponds to a non-degenerate ellipse in ${\mathbb{R}}^2$ whether $\mathrm{sgn}\left(A\right)=-\mathrm{sgn}(\Delta )\ne 0$. Then

(i)

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(R\right)\ne 0$, then Lemma 2(i) implies $-\mathrm{sgn}(\Delta )=\mathrm{sgn}\left(A\right)=\mathrm{sgn}(A+C)=\mathrm{sgn}\left(R\right)\ne 0$. Hence, consider the following complementary cases:

• Case 1. If $\mathrm{sgn}(A+C)=\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )=-\mathrm{sgn}\left(d\right)=1$, then ${\displaystyle \frac{\Delta }{d}}>0$. In addition, Lemma 3(i) guarantees $A+C>R>0$, so this and Remark 1 imply $A^{\prime }={\displaystyle \frac{A+C+R}{2}}>C^{\prime }={\displaystyle \frac{A+C-R}{2}}>0.$ Thus ${\displaystyle \frac{1}{C^{\prime }}}>{\displaystyle \frac{1}{A^{\prime }}}>0$, hence ${\displaystyle \frac{4\Delta }{C^{\prime }d}}>{\displaystyle \frac{4\Delta }{A^{\prime }d}}>0.$
• Case 2. If $\mathrm{sgn}(A+C)=\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )=\mathrm{sgn}\left(d\right)=-1$. Then ${\displaystyle \frac{\Delta }{d}}<0.$ In addition, Lemma 3(i) guarantees $A+C<R<0$, so this and Remark 1 imply $A^{\prime }={\displaystyle \frac{A+C+R}{2}}<C^{\prime }={\displaystyle \frac{A+C-R}{2}}<0$. Thus ${\displaystyle \frac{1}{C^{\prime }}}<{\displaystyle \frac{1}{A^{\prime }}}<0$, hence ${\displaystyle \frac{4\Delta }{C^{\prime }d}}>{\displaystyle \frac{4\Delta }{A^{\prime }d}}>0.$

Thereby, Remarks 1 and 2(b) imply $a^2={\displaystyle \frac{4\Delta }{C^{\prime }d}}={\displaystyle \frac{8\Delta }{(A+C-R)d}}>b^2={\displaystyle \frac{4\Delta }{A^{\prime }d}}={\displaystyle \frac{8\Delta }{(A+C+R)d}}>0$ for both cases, thus

$$a=2\sqrt{\frac{2\Delta }{(A+C-R)d}}>b=2\sqrt{\frac{2\Delta }{(A+C+R)d}}>0.$$

On the other hand, consider that $c=\sqrt{a^2-b^2}>0$ (see (Ref. [7], pp. 161–164). Therefore, Corollary 2(i) and Remark 8 imply

$$\begin{array}{ccc}\hfill c& =& \sqrt{\frac{8\Delta }{(A+C-R)d}-\frac{8\Delta }{(A+C+R)d}}=\sqrt{\frac{8\Delta }{d}\left(\frac{1}{A+C-R}-\frac{1}{A+C+R}\right)}=\sqrt{\frac{8\Delta }{d}\left[\frac{2R}{{(A+C)}^2-R^2}\right]}\hfill \\ & =& 4\sqrt{\frac{R\Delta }{d[{(A+C)}^2-{(A-C)}^2-B^2]}}=4\sqrt{\frac{R\Delta }{d[-(B^2-4AC)]}}=4\sqrt{-\frac{R\Delta }{d^2}}=-\frac{4\sqrt{-R\Delta }}{d}>0.\hfill \end{array}$$

(ii)

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$, then Lemma 2(i) implies $-\mathrm{sgn}(\Delta )=\mathrm{sgn}\left(A\right)=\mathrm{sgn}(A+C)=-\mathrm{sgn}\left(R\right)\ne 0$. Hence, consider the following complementary cases:

• Case 1. If $\mathrm{sgn}(A+C)=-\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )=-\mathrm{sgn}\left(d\right)=1$, then ${\displaystyle \frac{\Delta }{d}}>0$. In addition, Lemma 3(i) guarantees $A+C>-R>0>R$. So this fact, along with Remark 1 implies $C^{\prime }={\displaystyle \frac{A+C-R}{2}}>A^{\prime }={\displaystyle \frac{A+C+R}{2}}>0$. Thus ${\displaystyle \frac{1}{A^{\prime }}}>{\displaystyle \frac{1}{C^{\prime }}}>0$, hence ${\displaystyle \frac{4\Delta }{A^{\prime }d}}>{\displaystyle \frac{4\Delta }{C^{\prime }d}}>0$.
• Case 2. If $\mathrm{sgn}(A+C)=-\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )=\mathrm{sgn}\left(d\right)=-1$, then ${\displaystyle \frac{\Delta }{d}}<0$. In addition, Lemma 3(i) guarantees $A+C<-R<0<R$, so this and Remark 1 imply $C^{\prime }={\displaystyle \frac{A+C-R}{2}}<A^{\prime }={\displaystyle \frac{A+C+R}{2}}<0$. Thus ${\displaystyle \frac{1}{A^{\prime }}}<{\displaystyle \frac{1}{C^{\prime }}}<0$, hence ${\displaystyle \frac{4\Delta }{A^{\prime }d}}>{\displaystyle \frac{4\Delta }{C^{\prime }d}}>0.$

Additionally, Remarks 1 and 2(b) imply $a^2={\displaystyle \frac{4\Delta }{A^{\prime }d}}={\displaystyle \frac{8\Delta }{(A+C+R)d}}>b^2={\displaystyle \frac{4\Delta }{C^{\prime }d}}={\displaystyle \frac{8\Delta }{(A+C-R)d}}>0$ for both cases; thus,

$$a=2\sqrt{\frac{2\Delta }{(A+C+R)d}}>b=2\sqrt{\frac{2\Delta }{(A+C-R)d}}>0.$$

Finally, Corollary 2(i) and Remark 8 imply

$$c=\sqrt{a^2-b^2}=\sqrt{\frac{8\Delta }{(A+C+R)d}-\frac{8\Delta }{(A+C-R)d}}=\sqrt{-\frac{8\Delta }{d}\left(\frac{1}{A+C-R}-\frac{1}{A+C+R}\right)}=4\sqrt{-\left(-\frac{R\Delta }{d^2}\right)}=\frac{4\sqrt{R\Delta }}{\sqrt{d^2}}=-\frac{4\sqrt{R\Delta }}{d}>0.$$

This completes the proof. □

Now, we point out some interrelations between the conic radical, the coefficients of the general equation of the non-degenerate ellipse, and the conic discriminant.

Proposition 3.

Let $A,B,C\in \mathbb{R}$ be such that $d<0$. Then

(i) 

$|A+C\pm R|=|A+C|\pm \left|R\right|>0$ whether $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(R\right)\ne 0$.

(ii) 

$|A+C\pm R|=|A+C|\mp \left|R\right|>0$ whether $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$.

Proof. 

Note that $d<0$ guarantees $|A+C|-\left|R\right|>0$, according to Lemma 2(ii). So, $|A+C|+\left|R\right|>0$ as well, which guarantees $\left||A+C|\pm \left|R\right|\right|=|A+C|\pm \left|R\right|>0$ in general. Additionally, consider that $\mathrm{sgn}\left(A\right)\ne 0$ implies $\left|\mathrm{sgn}\right(A\left)\right|=|\pm 1|=1$. Therefore,

(i)

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(R\right)\ne 0$, then Lemma 2(i) implies $\mathrm{sgn}\left(A\right)=\mathrm{sgn}(A+C)=\mathrm{sgn}\left(R\right)\ne 0$. Thus, Property 1 implies

$$\begin{array}{ccc}\hfill |A+C\pm R|& =& \left|\mathrm{sgn}(A+C)|A+C|\pm \mathrm{sgn}\left(R\right)\left|R\right|\right|=\left|\mathrm{sgn}\left(A\right)|A+C|\pm \mathrm{sgn}\left(A\right)\left|R\right|\right|\hfill \\ & =& \left|\mathrm{sgn}\left(A\right)\left(\right|A+C|\pm |R\left|\right)\right|=\left|\mathrm{sgn}\left(A\right)\right|\left||A+C|\pm \left|R\right|\right|=|A+C|\pm \left|R\right|>0.\hfill \end{array}$$

(ii)

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$, then Lemma 2(i) implies $\mathrm{sgn}\left(A\right)=\mathrm{sgn}(A+C)=-\mathrm{sgn}\left(R\right)\ne 0$. Thus, Property 1 implies

$$\begin{array}{ccc}\hfill |A+C\pm R|& =& \left|\mathrm{sgn}(A+C)|A+C|\pm \mathrm{sgn}\left(R\right)\left|R\right|\right|=\left|\mathrm{sgn}\left(A\right)|A+C|\pm (-\mathrm{sgn}(A\left)\right)\left|R\right|\right|\hfill \\ & =& \left|\mathrm{sgn}\left(A\right)\left(\right|A+C|\mp |R\left|\right)\right|=\left|\mathrm{sgn}\left(A\right)\right|\left||A+C|\mp \left|R\right|\right|=|A+C|\mp \left|R\right|>0.\hfill \end{array}$$

This completes the result. □

Although Lemma 7 seems to give some general formulae for the lengths of the semi-major and semi-minor axes and for the least distance between the center of the ellipse and the foci (the couple of focal points of the ellipse), Proposition 3 helps to develop the subsequent results, which are even more general than Lemma 7.

Theorem 13

(General theorem of the elliptical semiaxes). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate ellipse on the Cartesian plane, so if a and b are, respectively, the lengths of its semi-major and semi-minor axes. Then, these ones are given by the following general formulae:

$$a=2\sqrt{\frac{2|\Delta |}{\left(\right|R|-|A+C\left|\right)d}}>0andb=2\sqrt{-\frac{2|\Delta |}{\left(\right|A+C|+|R\left|\right)d}}>0.$$

Proof. 

Consider the following complementary cases:

Case 1. If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(R\right)\ne 0$, then Lemma 7(i), Remark 8, and Proposition 3(i) imply

$$\begin{array}{ccc}\hfill a& =& 2\sqrt{\frac{2\Delta }{(A+C-R)d}}=2\sqrt{\left|\frac{2\Delta }{(A+C-R)d}\right|}=2\sqrt{\frac{2|\Delta |}{|A+C-R|\left|d\right|}}=2\sqrt{\frac{2|\Delta |}{\left(\right|A+C|-|R\left|\right)(-d)}}=2\sqrt{\frac{2|\Delta |}{\left(\right|R|-|A+C\left|\right)d}},\hfill \\ \hfill b& =& 2\sqrt{\frac{2\Delta }{(A+C+R)d}}=2\sqrt{\left|\frac{2\Delta }{(A+C+R)d}\right|}=2\sqrt{\frac{2|\Delta |}{|A+C+R|\left|d\right|}}=2\sqrt{\frac{2|\Delta |}{\left(\right|A+C|+|R\left|\right)(-d)}}=2\sqrt{-\frac{2|\Delta |}{\left(\right|A+C|+|R\left|\right)d}}>0.\hfill \end{array}$$

Case 2. If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$, then Lemma 7(ii), Remark 8, and Proposition 3(ii) imply

$$\begin{array}{ccc}\hfill a& =& 2\sqrt{\frac{2\Delta }{(A+C+R)d}}=2\sqrt{\left|\frac{2\Delta }{(A+C+R)d}\right|}=2\sqrt{\frac{2|\Delta |}{|A+C+R|\left|d\right|}}=2\sqrt{\frac{2|\Delta |}{\left(\right|A+C|-|R\left|\right)(-d)}}=2\sqrt{\frac{2|\Delta |}{\left(\right|R|-|A+C\left|\right)d}},\hfill \\ \hfill b& =& 2\sqrt{\frac{2\Delta }{(A+C-R)d}}=2\sqrt{\left|\frac{2\Delta }{(A+C-R)d}\right|}=2\sqrt{\frac{2|\Delta |}{|A+C-R|\left|d\right|}}=2\sqrt{\frac{2|\Delta |}{\left(\right|A+C|+|R\left|\right)(-d)}}=2\sqrt{-\frac{2|\Delta |}{\left(\right|A+C|+|R\left|\right)d}}>0.\hfill \end{array}$$

This completes the proof. □

Now we employ Theorem 13 to state a general formula for the least distance between the center and the foci of the non-degenerate ellipse.

Corollary 10

(The general formula for the least distance between the center and the foci of the ellipse). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate ellipse on the Cartesian plane. Then, the least distance between its central point and each of its two focal points is given by $c=-{\displaystyle \frac{4\sqrt{|R\Delta |}}{d}}>0.$

Proof. 

Theorem 13, (11) and Remark 8 imply

$$\begin{array}{ccc}\hfill c& =& \sqrt{a^2-b^2}=\sqrt{{\left[2\sqrt{\frac{2|\Delta |}{\left(\right|R|-|A+C\left|\right)d}}\right]}^2-{\left[2\sqrt{-\frac{2|\Delta |}{\left(\right|A+C|+|R\left|\right)d}}\right]}^2}=\sqrt{\frac{8|\Delta |}{\left(\right|R|-|A+C\left|\right)d}+\frac{8|\Delta |}{\left(\right|A+C|+|R\left|\right)d}}\hfill \\ & =& \sqrt{\frac{8|\Delta |}{d}\left(\frac{1}{\left|R\right|-|A+C|}+\frac{1}{\left|R\right|+|A+C|}\right)}=\sqrt{\frac{8|\Delta |}{d}\left[\frac{2\left|R\right|}{{\left|R\right|}^2-{(A+C)}^2}\right]}=4\sqrt{\frac{\left|R\right||\Delta |}{d(B^2-4AC)}}=-\frac{4\sqrt{|R\Delta |}}{d}>0.\hfill \end{array}$$

This completes the proof. □

Now we present another consequence of Theorem 13. It serves the purpose of stating the length of the latera recta of the non-degenerate ellipse.

Corollary 11

(The general formula for the length of the latera recta of the ellipse). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate ellipse on the Cartesian plane; then the length of its two latera recta is given by the following general formula:

$${\ell }_r=\frac{4}{|A+C|+\left|R\right|}\sqrt{\frac{2\left(\right|R|-|A+C\left|\right)|\Delta |}{d}}>0.$$

Proof. 

According to [Ref. [10], p. 211], the length of the two latera recta of a non-degenerate ellipse is ${\ell }_r={\displaystyle \frac{2b^2}{a}}>0$. Then, Theorem 13 and Remark 8 imply

$$\begin{array}{c}\hfill {\ell }_r={\displaystyle \frac{2\left[-\frac{8|\Delta |}{\left(\right|A+C|+|R\left|\right)d}\right]}{2\sqrt{\frac{2|\Delta |}{\left(\right|R|-|A+C\left|\right)d}}}}=\frac{8|\Delta |}{\left(\right|A+C|+|R\left|\right)(-d)}\sqrt{\frac{\left(\right|R|-|A+C\left|\right)d}{2|\Delta |}}=\frac{4\sqrt{{|\Delta |}^2}}{\left(\right|A+C|+|R\left|\right)\sqrt{d^2}}\sqrt{\frac{2\left(\right|R|-|A+C\left|\right)d}{|\Delta |}}=\frac{4}{|A+C|+\left|R\right|}\sqrt{\frac{2\left(\right|R|-|A+C\left|\right)|\Delta |}{d}}>0.\end{array}$$

This completes the proof. □

5.2. The Equation of the Focal Axis of the Ellipse

Our next couple of results gives the equation of the straight line that corresponds to the focal axis of the non-degenerate ellipse in terms of the coefficients of the latter and the corresponding conic discriminant and conic radical.

Corollary 12.

Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate ellipse on the Cartesian plane; then the equation of the straight line that corresponds to its focal axis is given as follows:

(i) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(R\right)\ne 0$, then the equation is $y=-\sqrt{\frac{A-C+R}{C-A+R}}x+\frac{\sqrt{A-C+R}(2CD-BE)+\sqrt{C-A+R}(2AE-BD)}{\sqrt{C-A+R}d}$, whenever $B\ne 0$, otherwise the equation is $x=-{\displaystyle \frac{D}{2A}}$.

(ii) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$, then the equation is $y=\sqrt{\frac{C-A+R}{A-C+R}}x+\frac{\sqrt{A-C+R}(2AE-BD)-\sqrt{C-A+R}(2CD-BE)}{\sqrt{A-C+R}d}$.

Proof. 

(i) According to Remarks 1 and 2(b) and Lemma 7(i), the semi-major axis a is on the $y^{″}$-axis of Figure 1. Hence, Theorem 7(ii) and Remark 5 yield the result in this case.

(ii) According to Remarks 1 and 2(b) and Lemma 7(ii), the semi-major axis a is on the $x^{″}$-axis of Figure 1. Hence, Theorem 7(i) and Remark 5 yield the result in this other case.

This completes the proof. □

The proof of the following result is similar to that of Corollary 12. The only difference that Corollary 4 is applied instead of Theorem 7, whereas (12) guarantees $\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(R\right)\ne 0$ whenever $\theta \ne 0\ne B$.

Corollary 13.

Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate ellipse on the Cartesian plane, rotated by an angle $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$. Then, the equation of the straight line that corresponds to its focal axis is given as follows:

(i) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)\ne 0$, then the equation is

$$y=\frac{B}{A-C-R}x+\frac{(A-C-R)(2AE-BD)-B(2CD-BE)}{(A-C-R)d}.$$

(ii) 

If $\mathrm{sgn}\left(A\right)=-\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)\ne 0$, then the equation is

$$y=\frac{C-A+R}{B}x+\frac{B(2AE-BD)-(C-A+R)(2CD-BE)}{Bd}.$$

It is clear that Corollary 12 is more general than Corollary 13 because Corollary 12 works for all the non-degenerate ellipses, rotated or not. Meanwhile, Corollary 13 works only for the rotated ellipses.

5.3. On the Extreme and Focal Points of the Ellipse

This subsection is dedicated to presenting all the general formulae for the coordinates of the extreme points (vertices and co-vertices) and the foci of any non-degenerate ellipse. We begin by giving a general theorem for the vertices of the ellipse in terms of the coefficients of its general equation, the conic discriminant, the conic radical, and the conic determinant.

Theorem 14

(General theorem of the vertices of the ellipse). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate ellipse on the Cartesian plane; then its vertices are given as follows:

(i) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(R\right)\ne 0$, then

$${\mathcal{V}}_{1,2}\left(\frac{2CD-BE}{d}\mp 2\sqrt{\frac{(C-A+R)\Delta }{(A+C-R)Rd}},\frac{2AE-BD}{d}\pm 2\sqrt{\frac{(A-C+R)\Delta }{(A+C-R)Rd}}\right).$$

(ii) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$, then

$${\mathcal{V}}_{1,2}\left(\frac{2CD-BE}{d}\pm 2\sqrt{\frac{(A-C+R)\Delta }{(A+C+R)Rd}},\frac{2AE-BD}{d}\pm 2\sqrt{\frac{(C-A+R)\Delta }{(A+C+R)Rd}}\right).$$

Proof. 

Suppose that $\mathcal{V}(x_{\mathcal{V}},y_{\mathcal{V}})$ is one of the vertices of the ellipse on the non-rotated Cartesian plane, which is determined by some point ${\mathcal{V}}^{\prime }$ on the rotated Cartesian plane. So, consider the following complementary cases:

(i)

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(R\right)\ne 0$, then Corollary 12 and Theorem 7 guarantee that the focal axis of the ellipse is vertical with respect to the rotated Cartesian plane, so ${\mathcal{V}}^{\prime }(h^{\prime },k^{\prime }\pm a)$, where $h^{\prime }$ and $k^{\prime }$ are the coordinates of the central point of the ellipse on the rotated Cartesian plane. Hence, (6)–(8), (15), (16), (18) and Lemma 7 imply

$$\begin{array}{ccc}\hfill x_{\mathcal{V}}& =& h^{\prime }cos\theta -(k^{\prime }\pm a)sin\theta =(hcos\theta +ksin\theta )cos\theta -(-hsin\theta +kcos\theta \pm a)sin\theta \hfill \\ & =& h({sin}^2\theta +{cos}^2\theta )+ksin\theta cos\theta -ksin\theta cos\theta \mp asin\theta =h\mp asin\theta \hfill \\ & =& \frac{2CD-BE}{B^2-4AC}\mp \left(2\sqrt{\frac{2\Delta }{(A+C-R)d}}\sqrt{\frac{C-A+R}{2R}}\right)=\frac{2CD-BE}{d}\mp 2\sqrt{\frac{(C-A+R)\Delta }{(A+C-R)Rd}},\hfill \\ \hfill y_{\mathcal{V}}& =& h^{\prime }sin\theta +(k^{\prime }\pm a)cos\theta =(hcos\theta +ksin\theta )sin\theta +(-hsin\theta +kcos\theta \pm a)cos\theta \hfill \\ & =& hsin\theta cos\theta +k({sin}^2\theta +{cos}^2\theta )-hsin\theta cos\theta \pm acos\theta =k\pm acos\theta \hfill \\ & =& \frac{2AE-BD}{B^2-4AC}\pm \left(2\sqrt{\frac{2\Delta }{(A+C-R)d}}\sqrt{\frac{A-C+R}{2R}}\right)=\frac{2AE-BD}{d}\pm 2\sqrt{\frac{(A-C+R)\Delta }{(A+C-R)Rd}}.\hfill \end{array}$$

(ii)

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$, then Corollary 12(ii) and Theorem 7(i) guarantee that the focal axis of the ellipse is horizontal with respect to the rotated Cartesian plane, so ${\mathcal{V}}^{\prime }(h^{\prime }\pm a,k^{\prime })$, where $h^{\prime }$ and $k^{\prime }$ are the coordinates of the center of the ellipse on the rotated Cartesian plane. Hence, (6)–(8), (15), (16), (18), and Lemma 7 imply

$$\begin{array}{ccc}\hfill x_{\mathcal{V}}& =& (h^{\prime }\pm a)cos\theta -k^{\prime }sin\theta =(hcos\theta +ksin\theta \pm a)cos\theta -(-hsin\theta +kcos\theta )sin\theta \hfill \\ & =& h({sin}^2\theta +{cos}^2\theta )+ksin\theta cos\theta \pm acos\theta -ksin\theta cos\theta =h\pm acos\theta \hfill \\ & =& \frac{2CD-BE}{B^2-4AC}\pm \left(2\sqrt{\frac{2\Delta }{(A+C+R)d}}\sqrt{\frac{A-C+R}{2R}}\right)=\frac{2CD-BE}{d}\pm 2\sqrt{\frac{(A-C+R)\Delta }{(A+C+R)Rd}},\hfill \\ \hfill y_{\mathcal{V}}& =& (h^{\prime }\pm a)sin\theta +k^{\prime }cos\theta =(hcos\theta +ksin\theta \pm a)sin\theta +(-hsin\theta +kcos\theta )cos\theta \hfill \\ & =& hsin\theta cos\theta \pm asin\theta +k({sin}^2\theta +{cos}^2\theta )-hsin\theta cos\theta =k\pm asin\theta \hfill \\ & =& \frac{2AE-BD}{B^2-4AC}\pm \left(2\sqrt{\frac{2\Delta }{(A+C+R)d}}\sqrt{\frac{C-A+R}{2R}}\right)=\frac{2AE-BD}{d}\pm 2\sqrt{\frac{(C-A+R)\Delta }{(A+C+R)Rd}}.\hfill \end{array}$$

This completes the proof. □

Now we exhibit a general theorem for the co-vertices of the ellipse in terms of the coefficients of its general equation, the conic discriminant, the conic radical, and the conic determinant.

Theorem 15

(General theorem of the co-vertices of the ellipse). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate ellipse on the Cartesian plane; then its co-vertices are given as follows:

(i) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(R\right)\ne 0$, then

$${\mathcal{B}}_{1,2}\left(\frac{2CD-BE}{d}\pm 2\sqrt{\frac{(A-C+R)\Delta }{(A+C+R)Rd}},\frac{2AE-BD}{d}\pm 2\sqrt{\frac{(C-A+R)\Delta }{(A+C+R)Rd}}\right).$$

(ii) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$, then

$${\mathcal{B}}_{1,2}\left(\frac{2CD-BE}{d}\mp 2\sqrt{\frac{(C-A+R)\Delta }{(A+C-R)Rd}},\frac{2AE-BD}{d}\pm 2\sqrt{\frac{(A-C+R)\Delta }{(A+C-R)Rd}}\right).$$

Proof. 

Suppose that $\mathcal{B}(x_{\mathcal{B}},y_{\mathcal{B}})$ is one of the co-vertices of the ellipse on the non-rotated Cartesian plane, which is determined by some point ${\mathcal{B}}^{\prime }$ on the rotated Cartesian plane. So, consider the following complementary cases:

(i)

Suppose that $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(R\right)\ne 0$, so if $h^{\prime }$ and $k^{\prime }$ are the coordinates of the central point of the ellipse on the rotated Cartesian plane, then Corollary 12(i) and Theorem 7(ii) guarantee that the focal axis of the ellipse is vertical with respect to the rotated Cartesian plane, thus ${\mathcal{B}}^{\prime }(h^{\prime }\pm b,k^{\prime })$. Hence, (6)–(8), (15), (16), (18), and Lemma 7(i) imply

$$\begin{array}{ccc}\hfill x_{\mathcal{B}}& =& (h^{\prime }\pm b)cos\theta -k^{\prime }sin\theta =(hcos\theta +ksin\theta \pm b)cos\theta -(-hsin\theta +kcos\theta )sin\theta \hfill \\ & =& h({sin}^2\theta +{cos}^2\theta )+ksin\theta cos\theta \pm bcos\theta -ksin\theta cos\theta =h\pm bcos\theta \hfill \\ & =& \frac{2CD-BE}{B^2-4AC}\pm 2\sqrt{\frac{2\Delta }{(A+C+R)d}}\sqrt{\frac{A-C+R}{2R}}=\frac{2CD-BE}{d}\pm 2\sqrt{\frac{(A-C+R)\Delta }{(A+C+R)Rd}},\hfill \\ \hfill y_{\mathcal{B}}& =& (h^{\prime }\pm b)sin\theta +k^{\prime }cos\theta =(hcos\theta +ksin\theta \pm b)sin\theta +(-hsin\theta +kcos\theta )cos\theta \hfill \\ & =& hsin\theta cos\theta +k({sin}^2\theta +{cos}^2\theta )\pm bsin\theta -ksin\theta cos\theta =k\pm bsin\theta \hfill \\ & =& \frac{2AE-BD}{B^2-4AC}\pm 2\sqrt{\frac{2\Delta }{(A+C+R)d}}\sqrt{\frac{C-A+R}{2R}}=\frac{2AE-BD}{d}\pm 2\sqrt{\frac{(C-A+R)\Delta }{(A+C+R)Rd}}.\hfill \end{array}$$

(ii)

Suppose that $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$, so if $h^{\prime }$ and $k^{\prime }$ are the coordinates of the central point of the ellipse on the rotated Cartesian plane, then Corollary 12(i) and Theorem 7(ii) guarantee that the focal axis of the ellipse is horizontal with respect to the rotated Cartesian plane, thus ${\mathcal{B}}^{\prime }(h^{\prime },k^{\prime }\pm b)$. Hence, (6)–(8), (15), (16), (18), and Lemma 7(ii) imply

$$\begin{array}{ccc}\hfill x_{\mathcal{B}}& =& h^{\prime }cos\theta -(k^{\prime }\pm b)sin\theta =(hcos\theta +ksin\theta )cos\theta -(-hsin\theta +kcos\theta \pm b)sin\theta \hfill \\ & =& h({sin}^2\theta +{cos}^2\theta )+ksin\theta cos\theta \mp bsin\theta -ksin\theta cos\theta =h\pm bsin\theta \hfill \\ & =& \frac{2CD-BE}{B^2-4AC}\pm 2\sqrt{\frac{2\Delta }{(A+C-R)d}}\sqrt{\frac{C-A+R}{2R}}=\frac{2CD-BE}{d}\pm 2\sqrt{\frac{(C-A+R)\Delta }{(A+C-R)Rd}},\hfill \\ \hfill y_{\mathcal{B}}& =& h^{\prime }sin\theta +(k^{\prime }\pm b)cos\theta =(hcos\theta +ksin\theta )sin\theta +(-hsin\theta +kcos\theta \pm b)cos\theta \hfill \\ & =& hsin\theta cos\theta -hsin\theta cos\theta +k({sin}^2\theta +{cos}^2\theta )\pm bcos\theta =k\pm bcos\theta \hfill \\ & =& \frac{2AE-BD}{B^2-4AC}\pm 2\sqrt{\frac{2\Delta }{(A+C-R)d}}\sqrt{\frac{A-C+R}{2R}}=\frac{2AE-BD}{d}\pm 2\sqrt{\frac{(A-C+R)\Delta }{(A+C-R)Rd}}.\hfill \end{array}$$

This completes the proof. □

Now we give a general theorem for the foci of the ellipse in terms of the coefficients of its general equation, the conic discriminant, the conic radical, and the conic determinant. Its proof is similar to that of Theorem 14, with the only difference that the expressions given by Lemma 7 for c are applied instead of the expressions for a given by the same lemma.

Theorem 16

(General theorem of the foci of the ellipse). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate ellipse on the Cartesian plane; then its focal points are given as follows:

(i) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(R\right)\ne 0$, then

$${\mathcal{F}}_{1,2}\left(\frac{2CD-BE\pm \sqrt{2(A-C-R)\Delta }}{d},\frac{2AE-BD\mp \sqrt{2(C-A-R)\Delta }}{d}\right).$$

(ii) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$, then

$${\mathcal{F}}_{1,2}\left(\frac{2CD-BE\mp \sqrt{2(A-C+R)\Delta }}{d},\frac{2AE-BD\mp \sqrt{2(C-A+R)\Delta }}{d}\right).$$

5.4. The Equations of the Directrices and the Latera Recta of the Ellipse

Remark 9.

According to ([6], pp. 107–108, 117–118), the least distance between the central point of an ellipse or a hyperbola and any of its two directrices is ${\delta }_D={\displaystyle \frac{a}{\epsilon }}$. So, if $\epsilon ={\displaystyle \frac{c}{a}}$ then it is clear that ${\delta }_D={\displaystyle \frac{a^2}{c}}$.

Now we give a formula for the least distance between the central point of the ellipse and any of its two directrices in terms of the coefficients of its general equation, the conic discriminant, the conic radical, and the conic determinant.

Lemma 8.

Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate ellipse on the Cartesian plane. Then, the least distance between the central point of this ellipse and any of its two directrices is given by the following general formula:

$${\delta }_D=\frac{2}{|A+C|-\left|R\right|}\sqrt{\left|\frac{\Delta }{R}\right|}>0.$$

Proof. 

It is clear that Theorem 13, Corollary 10, and Remark 9 imply

$$\begin{array}{c}\hfill {\delta }_D={\displaystyle \frac{a^2}{c}}=\frac{{\left(2\sqrt{\frac{2|\Delta |}{\left(\right|R|-|A+C\left|\right)d}}\right)}^2}{\left(-\frac{4\sqrt{|R\Delta |}}{d}\right)}=-\frac{2\sqrt{{|\Delta |}^2}d}{\left(\right|R|-|A+C\left|\right)\sqrt{\left|R\right||\Delta |}d}=\frac{2}{|A+C|-\left|R\right|}\sqrt{\left|\frac{\Delta }{R}\right|}>0.\end{array}$$

This yields the result. □

The following two results are simple observations regarding the forms of the terms involved in the definition of ${\delta }_D$.

Proposition 4.

Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate ellipse on the Cartesian plane. Then

(i) 

$\sqrt{\left|{\displaystyle \frac{\Delta }{R}}\right|}=\sqrt{-{\displaystyle \frac{\Delta }{R}}}>0$, whenever $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(R\right)\ne 0$,

(ii) 

$\sqrt{\left|{\displaystyle \frac{\Delta }{R}}\right|}=\sqrt{{\displaystyle \frac{\Delta }{R}}}>0$, whenever $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$.

Proof. 

Observe that the formulae given for c in Lemma 7 guarantee the following facts: $\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )\ne 0$ whether $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(R\right)\ne 0$, and $\mathrm{sgn}\left(R\right)=\mathrm{sgn}(\Delta )\ne 0$ whether $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$. Thus, these facts yield the result. □

Proposition 5.

Let $A,C\in \mathbb{R}$ be such that $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}(A-C)\ne 0$, then $|A+C|-|A-C|=2\left|A\right|$.

Proof. 

Case 1. If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}(A-C)=1$, then $A>0$, $C>0$, and $A-C<0$; thus, $|A+C|-|A-C|=A+C-[-(A-C\left)\right]=2A>0$. So, $2\left|A\right|=2A$.

Case 2. If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}(A-C)=-1$, then $A<0$, $C<0$, and $A-C>0$; thus, $|A+C|-|A-C|=-(A+C)-(A-C)=-2A>0$. So, $2\left|A\right|=-2A$. □

Now we give a general characterization of the equations of the directrices of the ellipse in terms of the coefficients of its general equation, the conic discriminant, the conic radical, and the conic determinant.

Theorem 17

(General characterization of the equations of the directrices of ellipses). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate ellipse on the Cartesian plane. Then, the equations of the straight lines that correspond to its two directrices are given as follows:

(i) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(R\right)\ne 0$, then the equations are

$$y=\sqrt{\frac{C-A+R}{A-C+R}}x+\left[\frac{\sqrt{A-C+R}(2AE-BD)-\sqrt{C-A+R}(2CD-BE)}{\sqrt{A-C+R}d}\pm \frac{2}{|A+C|-\left|R\right|}\sqrt{\frac{2\Delta }{C-A-R}}\right].$$

(ii) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$, then the equations are

$$\begin{array}{c}\hfill y=-\sqrt{\frac{A-C+R}{C-A+R}}x+\left[\frac{\sqrt{A-C+R}(2CD-BE)+\sqrt{C-A+R}(2AE-BD)}{\sqrt{C-A+R}d}\pm \frac{2}{|A+C|-\left|R\right|}\sqrt{\frac{2\Delta }{C-A+R}}\right],wheneverB\ne 0;\end{array}$$

otherwise, they are given by $x=-{\displaystyle \frac{1}{2A}}\left(D\mp \sqrt{{\displaystyle \frac{4ACF-A{E}^2-C{D}^2}{A-C}}}\right).$

Proof. 

(i) Corollary 12(i) implies that the two directrices of the ellipse are parallel to the $x^{″}$-axis of Figure 1, which is also in the middle of both directrices. Hence, Corollary 6, Remarks 4(i) and 5, Proposition 2(i), Lemma 8, and Proposition 3(i) imply that the equations of the directrices are given by $y=mx+\left(b\pm {\delta }_D\sqrt{m^2+1}\right)$, where

$$\begin{array}{c}\hfill m=\sqrt{\frac{C-A+R}{A-C+R}},b=\frac{\sqrt{(A-C+R)}(2AE-BD)-\sqrt{(C-A+R)}(2CD-BE)}{\sqrt{A-C+R}d}\mathrm{and}\\ \hfill {\delta }_D\sqrt{m^2+1}=\left(\frac{2}{|A+C|-\left|R\right|}\sqrt{\left|\frac{\Delta }{R}\right|}\right)\sqrt{\frac{2R}{A-C+R}}=\frac{2}{|A+C|-\left|R\right|}\sqrt{\left(-\frac{\Delta }{R}\right)\left(\frac{2R}{A-C+R}\right)}=\frac{2}{|A+C|-\left|R\right|}\sqrt{\frac{2\Delta }{C-A-R}}.\end{array}$$

(ii) Corollary 12(ii) implies that the two directrices of the ellipse are parallel to the $y^{″}$-axis of Figure 1, which is also in the middle of both directrices. Now consider the following two cases:

Case 1. Suppose that $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$, then $B\ne 0$. Hence, Corollary 6, Remarks 4(ii) and 5, Proposition 2(ii), Lemma 8, and Proposition 3(ii) imply that the equations of the directrices are given by $y=mx+\left(b\pm {\delta }_D\sqrt{m^2+1}\right)$, where

$$\begin{array}{c}\hfill m=-\sqrt{\frac{A-C+R}{C-A+R}},b=\frac{\sqrt{(A-C+R)}(2CD-BE)+\sqrt{(C-A+R)}(2AE-BD)}{\sqrt{C-A+R}d}\mathrm{and}\\ \hfill {\delta }_D\sqrt{m^2+1}=\left(\frac{2}{|A+C|-\left|R\right|}\sqrt{\left|\frac{\Delta }{R}\right|}\right)\sqrt{\frac{2R}{C-A+R}}=\frac{2}{|A+C|-\left|R\right|}\sqrt{\left(\frac{\Delta }{R}\right)\left(\frac{2R}{C-A+R}\right)}=\frac{2}{|A+C|-\left|R\right|}\sqrt{\frac{2\Delta }{C-A+R}}.\end{array}$$

Case 2. Suppose that $\theta =B=0$, then (10) guarantees $R=A-C$, whereas the equations of the two directrices are given by $x=h\pm {\displaystyle \frac{a}{\epsilon }}$ (see [6], p. 107). Hence, (5), (8), Remark 9, and Propositions 3(ii) and 4 imply

$$\begin{array}{ccc}\hfill x& =& h\pm \frac{a}{\epsilon }=\frac{2CD-BE}{B^2-4AC}\pm \frac{a^2}{c}=\frac{2CD-\left(0\right)E}{0^2-4AC}\pm {\delta }_D=-\frac{D}{2A}\pm \frac{2}{|A+C|-\left|R\right|}\sqrt{\frac{\Delta }{R}}\hfill \\ & =& -\frac{D}{2A}\pm \frac{2}{|A+C|-|A-C|}\sqrt{\frac{1}{A-C}\left[\frac{\left(0\right)DE-A{E}^2-C{D}^2-(0^2-4AC)F}{4}\right]}=-\frac{D}{2A}\pm \frac{1}{2\left|A\right|}\sqrt{\frac{4ACF-A{E}^2-C{D}^2}{A-C}}.\hfill \end{array}$$

Finally, observe that the absolute value becomes redundant in this expression due to the double sign, so this yields the result. □

The proof of the following result is similar to those of item (i) and case 1 of item (ii) of Theorem 17. The only difference is that Corollary 13 is used instead of Corollary 12.

Theorem 18

(Characterization of the Equations of the Directrices of Rotated Ellipses). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate ellipse on the Cartesian plane, rotated by an angle $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$. Then, the equations of the straight lines that correspond to its two directrices are given as follows:

(i) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)\ne 0$, then the equations are

$$y=\left(\frac{C-A+R}{B}\right)x+\left[\frac{B(2AE-BD)-(C-A+R)(2CD-BE)}{Bd}\pm \frac{2}{|A+C|-\left|R\right|}\sqrt{\frac{2\Delta }{C-A-R}}\right].$$

(ii) 

If $\mathrm{sgn}\left(A\right)=-\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)\ne 0$, then the equations are

$$y=\left(\frac{B}{A-C-R}\right)x+\left[\frac{(A-C-R)(2AE-BD)-B(2CD-BE)}{(A-C-R)d}\pm \frac{2}{|A+C|-\left|R\right|}\sqrt{\frac{2\Delta }{C-A+R}}\right].$$

Now we give a general characterization of the equations of the latera recta of ellipses in terms of the coefficients of its general equation, the conic discriminant, the conic radical, and the conic determinant. Its proof is similar to that of Theorem 17; the only difference is that the formulae given for c in Lemma 7 are applied instead of Lemma 8 and Proposition 3.

Theorem 19

(General characterization of the equations of the latera recta of ellipses). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate ellipse on the Cartesian plane. Then, the equations of the straight lines that correspond to its latera recta are given as follows:

(i) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(R\right)\ne 0$, then the equations are

$$y=\sqrt{\frac{C-A+R}{A-C+R}}x+\frac{1}{d}\left[\frac{\sqrt{A-C+R}(2AE-BD)-\sqrt{C-A+R}(2CD-BE)}{\sqrt{A-C+R}}\mp 4R\sqrt{\frac{2\Delta }{C-A-R}}\right].$$

(ii) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)\ne 0$, then the equations are

$$y=-\sqrt{\frac{A-C+R}{C-A+R}}x+\frac{1}{d}\left[\frac{\sqrt{A-C+R}(2CD-BE)+\sqrt{C-A+R}(2AE-BD)}{\sqrt{C-A+R}}\mp 4R\sqrt{\frac{2\Delta }{C-A+R}}\right],wheneverB\ne 0;$$

otherwise, they are given by $x=-{\displaystyle \frac{1}{2A}}\left(D\mp \sqrt{{\displaystyle \frac{(A-C)(4ACF-A{E}^2-C{D}^2)}{C}}}\right)$.

We conclude this subsection showing a result that states a general characterization of the equations of the latera recta of rotated ellipses in terms of the coefficients of its general equation, the conic discriminant, the conic radical, and the conic determinant. Its proof is straightforward, for it is similar to the proofs of item (i) and case 1 of item (ii) of Theorem 19. The only difference is that Corollary 13 is used instead of Corollary 12.

Theorem 20

(Characterization of the equations of the latera recta of rotated ellipses). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate ellipse on the Cartesian plane, rotated by an angle $\theta \in \left(0,{\displaystyle \frac{\pi }{2}}\right)$. Then, the equations of the straight lines that correspond to its latera recta are given as follows:

(i) 

If $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)\ne 0$, then the equations are

$$y=\left(\frac{C-A+R}{B}\right)x+\frac{1}{d}\left[\frac{B(2AE-BD)-(C-A+R)(2CD-BE)}{B}\mp 4R\sqrt{\frac{2\Delta }{C-A-R}}\right].$$

(ii) 

If $\mathrm{sgn}\left(A\right)=-\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(C\right)\ne 0$, then the equations are

$$y=\left(\frac{B}{A-C-R}\right)x+\frac{1}{d}\left[\frac{(A-C-R)(2AE-BD)-B(2CD-BE)}{A-C-R}\mp 4R\sqrt{\frac{2\Delta }{C-A+R}}\right].$$

It is clear that Theorem 17 is more general than Theorem 18, in the same sense of Remark 4 with respect to Theorem 6. Meanwhile, Theorem 19 is more general than Theorem 20 in the very same sense.

5.5. Examples of Non-Degenerate Ellipses

We devote this subsection to illustrate the theory we presented in this whole section. We begin, however, with a reminder of a fundamental property of the conic determinant in the case of non-degenerate ellipses.

Remark 10.

According to [2], (3) corresponds to a non-degenerate ellipse whenever $\mathrm{sgn}\left(A\right)=-\mathrm{sgn}(\Delta )\ne 0$, along with only one of the two following conditions: $A-C\ne B=0\ne \mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)$ for the non-rotated case and $B\ne 0>d$ for the rotated case. Therefore, it is clear that Theorem 3(ii) and Lemma 3 guarantee that the condition for the rotated case, which involves the discriminant of (3), is equivalent to $|A+C|>\left|R\right|>0\ne B$.

Example 3.

Suppose that $43x^2+14\sqrt{3}xy+57y^2-14\sqrt{3}x-114y-519=0$. Then $R=\mathrm{sgn}\left(14\sqrt{3}\right)\sqrt{{(43-57)}^2+{\left(14\sqrt{3}\right)}^2}=28$. Now observe that $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=\mathrm{sgn}\left(R\right)=1$, and $|A+C|=|43+57|=100>28=\left|R\right|>0\ne 14\sqrt{3}=B$. In addition, (5) implies $\Delta =-\mathrm{1,327,104}<0$, so $\mathrm{sgn}\left(A\right)=-\mathrm{sgn}(\Delta )=1$. Thereby, Remark 10 guarantees that the given equation corresponds to a non-degenerate ellipse, whose eccentricity is given by Theorem 5(i) as $\epsilon =\sqrt{{\displaystyle \frac{2\left|28\right|}{|43+57|+\left|28\right|}}}={\displaystyle \frac{\sqrt{7}}{4}}.$ Now consider that (18) gives $d=-9216$, thus Corollary 11 implies

$${\ell }_r=\frac{4}{|43+57|+\left|28\right|}\sqrt{\frac{2\left(\right|28|-|43+57\left|\right)-\left|\mathrm{1,327,104}\right|}{-9216}}=\frac{9}{2}.$$

Whereas Theorem 13 and Corollary 10 yield

$$\begin{array}{c}\hfill a=2\sqrt{\frac{2|-\mathrm{1,317,104}|}{\left(\right|28|-|43+57\left|\right)(-9216)}}=4,b=2\sqrt{-\frac{2|-\mathrm{1,327,104}|}{\left(\right|43+57|+|28\left|\right)(-9216)}}=3andc=-\frac{4\sqrt{\left|28\right(-\mathrm{1,327,104}\left)\right|}}{(-9216)}=\sqrt{7}.\end{array}$$

Additionally, (8) implies $h=0$ and $k=1$. So, the center of the ellipse is $\mathcal{C}(0,1)$. Hence, Theorem 14(i) implies that the two vertices of the ellipse are given by

$${\mathcal{V}}_{1,2}\left(0\mp 2\sqrt{\frac{(57-43+28)(-\mathrm{1,327,104})}{(43+57-28)\left(28\right)(-9216)}},1\pm 2\sqrt{\frac{(43-57+28)(-\mathrm{1,327,104})}{(43+57-28)\left(28\right)(-9216)}}\right).$$

Thus ${\mathcal{V}}_1(-2\sqrt{3},3)$ and ${\mathcal{V}}_2(2\sqrt{3},-1)$. Meanwhile, Theorem 15(i) implies that the two co-vertices of the ellipse are given by

$${\mathcal{B}}_{1,2}\left(0\pm 2\sqrt{\frac{(43-57+28)(-\mathrm{1,327,104})}{(43+57+28)\left(28\right)(-9216)}},1\pm 2\sqrt{\frac{(57-43+28)(-\mathrm{1,327,104})}{(43+57+28)\left(28\right)(-9216)}}\right).$$

So ${\mathcal{B}}_1\left({\displaystyle \frac{3}{2}},{\displaystyle \frac{2+3\sqrt{3}}{2}}\right)$ and ${\mathcal{B}}_2\left(-{\displaystyle \frac{3}{2}},{\displaystyle \frac{2-3\sqrt{3}}{2}}\right)$, and Theorem 16(i) implies that the two foci of the ellipse are given by

$${\mathcal{F}}_{1,2}\left(0\pm \frac{2\sqrt{2(43-57-28)(-\mathrm{1,327,104})}}{(-9216)},2\mp \frac{2\sqrt{2(57-43-28)(-\mathrm{1,327,104})}}{(-9216)}\right).$$

Hence ${\mathcal{F}}_1\left(-{\displaystyle \frac{\sqrt{21}}{2}},{\displaystyle \frac{2+\sqrt{7}}{2}}\right)$ and ${\mathcal{F}}_2\left({\displaystyle \frac{\sqrt{21}}{2}},{\displaystyle \frac{2-\sqrt{7}}{2}}\right)$. On the other hand, Corollary 12(i) guarantees that the equation of the focal axis of the ellipse is given by

$$y=-\sqrt{\frac{43-57+28}{57-43+28}}x+\frac{\sqrt{43-57+28}\left[2\left(57\right)(-14\sqrt{3})-\left(14\sqrt{3}\right)(-114)\right]+\sqrt{57-43+28}\left[2\left(43\right)(-114)-\left(14\sqrt{3}\right)(-14\sqrt{3})\right]}{\sqrt{57-43+28}(-9216)};$$

which is reduced to $y=-{\displaystyle \frac{1}{\sqrt{3}}}x+1$. In addition, Theorems 17(i) and 19(i) imply that the equations of the two directrices and the latera recta of the ellipse are respectively given by

$$y=\sqrt{\frac{57-43+28}{43-57+28}}x+\left[\frac{\sqrt{43-57+28}\left[2\left(43\right)(-114)-\left(14\sqrt{3}\right)(-14\sqrt{3})\right]-\sqrt{57-43+28}\left[2\left(57\right)(-14\sqrt{3})-\left(14\sqrt{3}\right)(-114)\right]}{\sqrt{43-57+28}(-9216)}\pm \frac{2}{|43+57|-\left|28\right|}\sqrt{\frac{2(-\mathrm{1,327,104})}{57-43-28}}\right],$$

$$y=\sqrt{\frac{57-43+28}{43-57+28}}x+\frac{1}{(-9216)}\left[\frac{\sqrt{43-57+28}\left[2\left(43\right)(-114)-\left(14\sqrt{3}\right)(-14\sqrt{3})\right]-\sqrt{57-43+28}\left[2\left(57\right)(-14\sqrt{3})-\left(14\sqrt{3}\right)(-114)\right]}{\sqrt{43-57+28}}\pm 4\left(28\right)\sqrt{\frac{2(-\mathrm{1,327,104})}{57-43-28}}\right],$$

which are respectively reduced to $y=\sqrt{3}x+{\displaystyle \frac{\sqrt{7}\pm 32}{\sqrt{7}}}$ and $y=\sqrt{3}x+\left(1\pm 2\sqrt{7}\right)$. Note that these equations can also be respectively obtained by Theorems 18(i) and 20(i). So, Corollary 5 implies that the general equations of the two degenerate cases of parabolas that respectively correspond to the directrices and the latera recta of the ellipse in question are $21x^2-14\sqrt{3}xy+7y^2+14\sqrt{3}x-14y-1017=0$ and $3x^2-2\sqrt{3}xy+y^2+2\sqrt{3}x-2y-27=0$.

Finally, note that the rotation angle was completely unneeded to obtain all these results. However, if it is required, then it can be obtained by applying Corollary 3 as $tan\theta =\sqrt{{\displaystyle \frac{57-43+28}{43-57+28}}}=\sqrt{3}$, thus $\theta =arctan\left(\sqrt{3}\right)=60°$. These geometric locations can be visualized in Figure 7.

Example 4.

If $1924x^2-1536xy+1476y^2+3080x-60y-\mathrm{20,975}=0$. Then $R=\mathrm{sgn}(-1536)\sqrt{{(1924-1476)}^2+{(-1536)}^2}=-1600$. Now observe that $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)=1$, and $|A+C|=|1924+1476|=3400>1600=\left|R\right|>0\ne -1536=B$. In addition, (5) implies $\Delta =-\mathrm{50,625,000,000}$, so $\mathrm{sgn}\left(A\right)=-\mathrm{sgn}(\Delta )=1$. Thereby, Remark 10 guarantees that the given equation corresponds to a non-degenerate ellipse, whose eccentricity is given by Theorem5(i) as follows: $\epsilon =\sqrt{{\displaystyle \frac{2|-1600|}{|1924+1476|+|-1600|}}}={\displaystyle \frac{4}{5}}$. Now consider that (18) gives $d=-\mathrm{9,000,000}$. Thus, Corollary 11 implies

$${\ell }_r=\frac{4}{|1924+1476|+|-1600|}\sqrt{\frac{2\left(\right|-1600|-|1924+1476\left|\right)|-\mathrm{50,625,000,000}|}{(-\mathrm{9,000,000})}}=\frac{18}{5}.$$

Whereas Theorem 13 and Corollary 10 yield

$$a=2\sqrt{\frac{2|-\mathrm{50,625,000,000}|}{\left(\right|-1600|-|1924+1476\left|\right)(-\mathrm{9,000,000})}}=5,b=2\sqrt{\frac{2|-\mathrm{50,625,000,000}|}{\left(\right|1924+1476|+|-1600\left|\right)(-\mathrm{9,000,000})}}=3$$

$$andc=-\frac{4\sqrt{|-1600(-\mathrm{50,625,000,000}\left)\right|}}{(-\mathrm{9,000,000})}=4.$$

Additionally, (8) implies $h=-1$ and $k=-{\displaystyle \frac{1}{2}}$, so the center of the ellipse is $\mathcal{C}(-1,-{\displaystyle \frac{1}{2}})$. Hence, Theorem 14(ii) implies that the two vertices of the ellipse are given by

$${\mathcal{V}}_{1,2}\left(-1\pm 2\sqrt{\frac{[1924-1476+(-1600\left)\right](-\mathrm{50,625,000,000})}{[1924+1476+(-1600\left)\right](-1600)(-\mathrm{9,000,000})}},-\frac{1}{2}\pm 2\sqrt{\frac{[1476-1924+(-1600\left)\right](-\mathrm{50,625,000,000})}{[1924+1476+(-1600\left)\right](-1600)(-\mathrm{9,000,000})}}\right).$$

Thus ${\mathcal{V}}_1(2,{\displaystyle \frac{7}{2}})$ and ${\mathcal{V}}_2(-4,-{\displaystyle \frac{9}{2}}).$ Meanwhile, Theorem 15(ii) implies that the two co-vertices of the ellipse are given by

$${\mathcal{B}}_{1,2}\left(-1\mp 2\sqrt{\frac{[1476-1924+(-1600\left)\right](-\mathrm{50,625,000,000})}{[1924+1476-(-1600\left)\right](-1600)(-\mathrm{9,000,000})}},-\frac{1}{2}\pm 2\sqrt{\frac{[1924-1476+(-1600\left)\right](-\mathrm{50,625,000,000})}{[1924+1476-(-1600\left)\right](-1600)(-\mathrm{9,000,000})}}\right).$$

So ${\mathcal{B}}_1\left(-{\displaystyle \frac{17}{5}},{\displaystyle \frac{13}{10}}\right)$ and ${\mathcal{B}}_2\left({\displaystyle \frac{7}{5}},-{\displaystyle \frac{23}{10}}\right)$. Theorem 16(ii) implies that the two foci of the ellipse are given by

$${\mathcal{F}}_1\left(-1\mp \frac{2\sqrt{2[1924-1476+(-1600\left)\right](-\mathrm{50,625,000,000})}}{-\mathrm{9,000,000}},-\frac{1}{2}\mp \frac{2\sqrt{2[1476-1924+(-1600\left)\right](-\mathrm{50,625,000,000})}}{-\mathrm{9,000,000}}\right).$$

Hence ${\mathcal{F}}_1\left({\displaystyle \frac{7}{5}},{\displaystyle \frac{27}{10}}\right)$ and ${\mathcal{F}}_2\left(-{\displaystyle \frac{17}{5}},-{\displaystyle \frac{37}{10}}\right)$. On the other hand, Corollary 13(ii) guarantees that the equation of the focal axis of the ellipse is given by

$$y=\left[\frac{1476-1924+(-1600)}{(-1536)}\right]x+\frac{(-1536)\left[2\right(1924\left)\right(-60)-(-1536\left)\right(3080\left)\right]-[1476-1924+(-1600\left)\right]\left[2\right(1476\left)\right(3080)-(-1536\left)\right(-60\left)\right]}{(-1536)(-\mathrm{9,000,000})}$$

which is reduced to $y={\displaystyle \frac{4}{3}}x+{\displaystyle \frac{5}{6}}$. In addition, Theorem 18(ii) implies that the equations of the two directrices of the ellipse are given by

$$y=\left[\frac{(-1536)}{1924-1476-(-1600)}\right]x+$$

$$\left[\frac{\left(1924-1476-(-1600)\right)\left(2\left(1924\right)(-60)-(-1536)\left(3080\right)\right)-(-1536)\left(2\left(1476\right)\left(3080\right)-(-1536)(-60)\right)}{\left(1924-1476-(-1600)\right)(-\mathrm{9,000,000})}\pm \frac{2}{|1924+1476|-|-1600|}\sqrt{\frac{2(-\mathrm{50,625,000,000})}{1476-1924+(-1600)}}\right],$$

which is reduced to $y=-{\displaystyle \frac{3}{4}}x+{\displaystyle \frac{105}{16}}$ and $y=-{\displaystyle \frac{3}{4}}x-{\displaystyle \frac{145}{16}}$. Meanwhile, Theorem 20(ii) implies that the equations of the latera recta of the ellipse are given by

$$y=\left[\frac{(-1536)}{1924-1476-(-1600)}\right]x+$$

$$\frac{1}{(-\mathrm{9,000,000})}\left[\frac{\left(1924-1476-(-1600)\right)\left(2\left(1924\right)(-60)-(-1536)\left(3080\right)\right)-(-1536)(2\left(1476\right)\left(3080\right)-(-1536)(-60))}{1924-1476-(-1600)}\pm 4(-1600)\sqrt{\frac{2(-\mathrm{50,625,000,000})}{1476-1924+(-1600)}}\right],$$

which is reduced to $y=-{\displaystyle \frac{3}{4}}x+{\displaystyle \frac{15}{4}}$ and $y=-{\displaystyle \frac{3}{4}}x-{\displaystyle \frac{25}{4}}$. Note that all the previous equations can also be respectively obtained by Theorems 17(ii) and 19(ii). So, Corollary 5 implies that the general equations of the two degenerate cases of parabolas that respectively correspond to the directrices and the latera recta of the ellipse in question are $144x^2+384xy+256y^2+480x+640y-\mathrm{15,225}=0$ and $9x^2+24xy+16y^2+30x+40y-375=0$.

Finally, note that the rotation angle was completely unneeded to obtain all these results. However, if it is required, then it can be obtained by applying Corollary 3 as $tan\theta =\sqrt{{\displaystyle \frac{1476-1924+(-1600)}{1924-1476+(-1600)}}}={\displaystyle \frac{4}{3}}$, thus $\theta =arctan\left({\displaystyle \frac{4}{3}}\right)\approx 53.13°$. All these geometric locations can be visualized in Figure 8.

Example 5.

Suppose that $-4x^2-9y^2-16x+18y+11=0$, then $R=A-C=-4-\left(-9\right)=5\ne 0=B$. Now observe that $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-\mathrm{sgn}\left(R\right)=-1$, and $|A+C|=\left|-4+(-9)\right|=13>5=\left|R\right|>0.$ In addition, (5) implies $\Delta =1296$, so $\mathrm{sgn}\left(A\right)=-\mathrm{sgn}(\Delta )=-1$. Thereby, Remark 10 guarantees that the given equation corresponds to a non-degenerate ellipse, whose eccentricity is given by Theorem 5(i) as $\epsilon =\sqrt{{\displaystyle \frac{2\left|5\right|}{|-4+(-9\left)\right|+\left|5\right|}}}={\displaystyle \frac{\sqrt{5}}{3}}.$ Now consider that (18) gives $d=-144$. Thus, Corollary 11 implies

$${\ell }_r=\frac{4}{|-4+(-9\left)\right|+\left|5\right|}\sqrt{\frac{2\left(\left|5\right|-|-4+(-9\left)\right|\right)\left|1296\right|}{(-144)}}=\frac{8}{3}.$$

Whereas Theorem 13 and Corollary 10 yield

$$a=2\sqrt{\frac{2\left|1296\right|}{\left(\left|5\right|-|-4+(-9\left)\right|\right)(-144)}}=3,b=2\sqrt{-\frac{2\left|1296\right|}{\left(|-4+(-9\left)\right|+\left|5\right|\right)(-144)}}=2andc=-\frac{4\sqrt{\left|5\right(1296\left)\right|}}{(-144)}=\sqrt{5}.$$

Additionally, (8) implies $h=-2$ and $k=1$, so the center of the ellipse is $\mathcal{C}(-2,1)$. Hence, Theorem 14(ii) implies that the two vertices of the ellipse are given by

$${\mathcal{V}}_{1,2}\left(-2\pm 2\sqrt{\frac{[-4-(-9)+5]\left(1296\right)}{[-4+(-9)+5]\left(5\right)(-144)}},1\pm 2\sqrt{\frac{[-9-(-4)+5]\left(1296\right)}{[-4+(-9)+5]\left(5\right)(-144)}}\right).$$

Thus ${\mathcal{V}}_1(1,1)$ and ${\mathcal{V}}_2(-5,1)$. Meanwhile, Theorem 15(ii) implies that the two co-vertices of the ellipse are given by

$${\mathcal{B}}_{1,2}\left(-2\mp 2\sqrt{\frac{[-9-(-4)+5]\left(1296\right)}{[-4+(-9)-5]\left(5\right)(-144)}},1\pm 2\sqrt{\frac{[-4-(-9)+5]\left(1296\right)}{[-4+(-9)-5]\left(5\right)(-144)}}\right).$$

So ${\mathcal{B}}_1(-2,3)$ and ${\mathcal{B}}_2(-2,-1)$, and Theorem 16(ii) implies that the two foci of the ellipse are given by

$${\mathcal{F}}_{1,2}\left(-2\mp \frac{2\sqrt{2[-4-(-9)+5]\left(1296\right)}}{(-144)},1\mp \frac{2\sqrt{2[-9-(-4)+5]\left(1296\right)}}{(-144)}\right).$$

Hence ${\mathcal{F}}_1\left(-2+\sqrt{5},1\right)$ and ${\mathcal{F}}_2\left(-2-\sqrt{5},1\right).$ On the other hand, Corollary 12(ii) guarantees that the equation of the focal axis of the ellipse is given by

$$\begin{array}{c}\hfill y=\sqrt{\frac{-9-(-4)+5}{-4-(-9)+5}}x+\frac{\sqrt{-4-(-9)+5}\left[2(-4)\left(18\right)-\left(0\right)(-16)\right]-\sqrt{-9-(-4)+5}\left[2(-9)(-16)-\left(0\right)\left(18\right)\right]}{\sqrt{-4-(-9)+5}(-144)};\end{array}$$

which is reduced to $y=1$. In addition, Theorems 17(ii) and 19(ii) imply that the equations of the two directrices and the latera recta of the ellipse are respectively given by

$$\begin{array}{ccc}\hfill x& =& -\frac{1}{2(-4)}\left[-16\pm \sqrt{\frac{4(-4)(-9)\left(11\right)-(-4){\left(18\right)}^2-(-9){(-16)}^2}{[-4-(-9\left)\right]}}\right]and\hfill \\ \hfill x& =& -\frac{1}{2(-4)}\left[-16\pm \frac{\sqrt{[-4-(-9)][4(-4)(-9)\left(11\right)-(-4){\left(18\right)}^2-(-9){(-16)}^2]}}{(-9)}\right],\hfill \end{array}$$

which are respectively reduced to $x=-2\pm {\displaystyle \frac{9}{\sqrt{5}}}$ and $x=-2\pm \sqrt{5}$. Note that in this case Theorems 18 and 20 cannot be applied. So, Corollary 5 implies that the general equations of the two degenerate cases of parabolas that respectively correspond to the directrices and the latera recta of the ellipse in question are $5x^2+20x-61=0$ and $x^2+4x-1=0.$ Finally, note that in this case Corollary 3 implies $tan\theta =\sqrt{{\displaystyle \frac{-9-(-4)+5}{-4-(-9)+5}}}=0$. So, $\theta =0°,$ which is totally congruent with the obvious fact that the given equation corresponds to a non-rotated ellipse. See Figure 9.

6. The Non-Degenerate Hyperbolas

This section is devoted to a thorough study of non-degenerate hyperbolas. We will present formulae for the lengths of semiaxes and latera recta, some particularities of equilateral hyperbolas, the equation of the focal axis, the extreme and focal points, the equations of the directrices and the latera recta, and the equations of the asymptotes. As before, we conclude the section with several numeric illustrations.

6.1. On the Lengths of the Semiaxes and the Latera Recta of the Hyperbola

We begin by presenting formulae for the lengths of the semiaxes and the minimum distance between the center of the locus and each focal point in terms of the coefficients of the general equation of the hyperbola, the conic discriminant, the conic radical, and the conic determinant.

Theorem 21.

Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate hyperbola on the Cartesian plane. If a and b are, respectively, the lengths of its two semiaxes, and c is the least distance between the center of the hyperbola and each of its two focal points, Then

(i) 

$a=2\sqrt{\frac{2\Delta }{(A+C+R)d}}>0$, $b=2\sqrt{\frac{2\Delta }{(R-A-C)d}}>0$ and $c=\frac{4\sqrt{R\Delta }}{d}>0$ whenever $\mathrm{sgn}\left(R\right)=\mathrm{sgn}(\Delta )\ne 0$.

(ii) 

$a=2\sqrt{\frac{2\Delta }{(A+C-R)d}}>0$, $b=2\sqrt{-\frac{2\Delta }{(A+C+R)d}}>0$ and $c=\frac{4\sqrt{-R\Delta }}{d}>0$ whenever $\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )\ne 0$.

Proof. 

According to [2], (3) corresponds to a non-degenerate hyperbola in ${\mathbb{R}}^2$ whenever $d>0\ne \Delta$. In addition, consider that $a^2+b^2=c^2>0$ (see (Ref. [10], p. 211), and observe that Remarks 1 and 2(b) guarantee the following two complementary cases:

(i)

Suppose that $a^2={\displaystyle \frac{4\Delta }{A^{\prime }d}}={\displaystyle \frac{8\Delta }{(A+C+R)d}}>0>-b^2={\displaystyle \frac{4\Delta }{C^{\prime }d}}={\displaystyle \frac{8\Delta }{(A+C-R)d}}$, then

$$a=\sqrt{\frac{8\Delta }{(A+C+R)d}}=2\sqrt{\frac{2\Delta }{(A+C+R)d}}>0\mathrm{and}b=\sqrt{-\frac{8\Delta }{(A+C-R)d}}=2\sqrt{\frac{2\Delta }{(R-A-C)d}}>0.$$

Hence, Remark 1 and Corollary 1(iii) imply

$$\begin{array}{ccc}\hfill c& =& \sqrt{a^2+b^2}=\sqrt{\frac{4\Delta }{A^{\prime }d}+\left(-\frac{4\Delta }{C^{\prime }d}\right)}=\sqrt{\left(\frac{4\Delta }{d}\right)\left(\frac{1}{A^{\prime }}-\frac{1}{C^{\prime }}\right)}=\sqrt{\left(\frac{16\Delta }{d}\right)\left(\frac{A^{\prime }-C^{\prime }}{-4A^{\prime }{C}^{\prime }}\right)}\hfill \\ & =& 4\sqrt{\left(\frac{\Delta }{d}\right)\left(\frac{\frac{A+C+R}{2}-\frac{A+C-R}{2}}{B^2-4AC}\right)}=4\sqrt{\left(\frac{\Delta }{d}\right)\left(\frac{R}{d}\right)}=4\sqrt{\frac{R\Delta }{d^2}}=\frac{4\sqrt{R\Delta }}{d}>0.\hfill \end{array}$$

Finally, note that this holds whenever $\mathrm{sgn}\left(R\right)=\mathrm{sgn}(\Delta )\ne 0$.

(ii)

Suppose that $a^2={\displaystyle \frac{4\Delta }{C^{\prime }d}}={\displaystyle \frac{8\Delta }{(A+C-R)d}}>0>-b^2={\displaystyle \frac{4\Delta }{A^{\prime }d}}={\displaystyle \frac{8\Delta }{(A+C+R)d}}$, then

$$a=\sqrt{\frac{8\Delta }{(A+C-R)d}}=2\sqrt{\frac{2\Delta }{(A+C-R)d}}>0\mathrm{and}b=\sqrt{-\frac{8\Delta }{(A+C+R)d}}=2\sqrt{-\frac{2\Delta }{(A+C+R)d}}>0.$$

Hence, item (i) implies

$$\begin{array}{c}\hfill c=\sqrt{a^2+b^2}=\sqrt{\frac{4\Delta }{C^{\prime }d}+\left(-\frac{4\Delta }{A^{\prime }d}\right)}=\sqrt{-\left(\frac{4\Delta }{d}\right)\left(\frac{1}{A^{\prime }}-\frac{1}{C^{\prime }}\right)}=4\sqrt{-\frac{R\Delta }{d^2}}=\frac{4\sqrt{-R\Delta }}{d}>0.\end{array}$$

Finally, note that this holds if $\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )\ne 0$.

This completes the proof. □

Now we state an important connection between the coefficients of the general equation of the hyperbola, the conic radical, and the conic determinant. It follows from the fact that, if $\Delta \ne 0$, then $\mathrm{sgn}(\Delta )=\pm 1$. So, $\left|R\right|>|A+C|$ guarantees $\left|R\right|>\pm (A+C)=\mathrm{sgn}(\Delta )(A+C)>-\left|R\right|$, thus $\mathrm{sgn}(\Delta )(A+C)+\left|R\right|>0$ and $-\mathrm{sgn}(\Delta )(A+C)+\left|R\right|>0$.

Proposition 6.

Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $\left|R\right|>|A+C|$ and $\Delta \ne 0$. Then

$$\left|\mathrm{sgn}(\Delta )(A+C)\pm \left|R\right|\right|=\left|\pm \mathrm{sgn}(\Delta )(A+C)+\left|R\right|\right|=\pm \mathrm{sgn}(\Delta )(A+C)+\left|R\right|>0.$$

Now we show general formulae for the lengths of the semiaxes of the hyperbola in terms of the coefficients of the general equation of the hyperbola, the conic discriminant, the conic radical, the conic discriminant and the conic determinant.

Corollary 14

(The general formulae for the lengths of the three semiaxes of the hyperbola). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate hyperbola on the Cartesian plane. Then, its three semiaxes a, b, and c are given by the following general formulae:

$$\begin{array}{c}\hfill a=2\sqrt{\frac{2|\Delta |}{\left[\mathrm{sgn}(\Delta )(A+C)+\left|R\right|\right]d}}>0,b=2\sqrt{\frac{2|\Delta |}{\left[-\mathrm{sgn}(\Delta )(A+C)+\left|R\right|\right]d}}>0andc=\frac{4\sqrt{|R\Delta |}}{d}>0.\end{array}$$

Proof. 

Firstly, define $s:=\mathrm{sgn}\left(R\right)\mathrm{sgn}(\Delta )$ and note that $\pm \mathrm{sgn}(\Delta )\ne 0$ and Property 3 guarantee ${\mathrm{sgn}}^2(\Delta )=1$, now consider the following complementary cases:

Case 1. If $\mathrm{sgn}\left(R\right)=\mathrm{sgn}(\Delta )\ne 0$, then $s=1$, so Theorem 21(i) and Property 2 imply

$$a=2\sqrt{\frac{2\Delta }{(A+C+R)d}}=2\sqrt{\frac{2\Delta }{(A+C+sR)d}·\frac{\mathrm{sgn}(\Delta )}{\mathrm{sgn}(\Delta )}}=2\sqrt{\frac{2\Delta \mathrm{sgn}(\Delta )}{\left[\mathrm{sgn}(\Delta )(A+C)+\left(R\mathrm{sgn}\left(R\right)\right){\mathrm{sgn}}^2(\Delta )\right]d}}=2\sqrt{\frac{2|\Delta |}{\left[\mathrm{sgn}(\Delta )(A+C)+\left|R\right|\right]d}}>0,$$

$$b=2\sqrt{\frac{2\Delta }{(R-A-C)d}}=2\sqrt{\frac{2\Delta }{[-(A+C)+sR]d}·\frac{\mathrm{sgn}(\Delta )}{\mathrm{sgn}(\Delta )}}=2\sqrt{\frac{2\Delta \mathrm{sgn}(\Delta )}{\left[-\mathrm{sgn}(\Delta )(A+C)+\left(R\mathrm{sgn}\left(R\right)\right){\mathrm{sgn}}^2(\Delta )\right]d}}=2\sqrt{\frac{2|\Delta |}{\left[-\mathrm{sgn}(\Delta )(A+C)+\left|R\right|\right]d}}>0.$$

Case 2. If $\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )\ne 0$, then $s=-1$, so Theorem 21(ii) and case 1 imply

$$\begin{array}{ccc}\hfill a& =& 2\sqrt{\frac{2\Delta }{(A+C-R)d}}=2\sqrt{\frac{2\Delta }{(A+C+sR)d}·\frac{\mathrm{sgn}(\Delta )}{\mathrm{sgn}(\Delta )}}=2\sqrt{\frac{2|\Delta |}{\left[\mathrm{sgn}(\Delta )(A+C)+\left|R\right|\right]d}}>0,\hfill \\ \hfill b& =& 2\sqrt{\frac{-2\Delta }{(A+C+R)d}}=2\sqrt{\frac{2\Delta }{[-(A+C)+sR)]d}·\frac{\mathrm{sgn}(\Delta )}{\mathrm{sgn}(\Delta )}}=2\sqrt{\frac{2|\Delta |}{\left[-\mathrm{sgn}(\Delta )(A+C)+\left|R\right|\right]d}}>0.\hfill \end{array}$$

Note that Lemma 3(iii) and Proposition 6 guarantee that all the formulae obtained in cases 1 and 2 always give real numbers. Finally, Theorem 21 guarantees

$$c=\frac{4\sqrt{R\Delta }}{d}>0\mathrm{whenever}\mathrm{sgn}\left(R\right)=\mathrm{sgn}(\Delta )\ne 0,\mathrm{so}|R\Delta |=R\Delta >0,$$

$$\mathrm{and}c=\frac{4\sqrt{-R\Delta }}{d}>0\mathrm{whenever}\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )\ne 0,\mathrm{so}|R\Delta |=-R\Delta >0.$$

Therefore, these facts yield the result. □

Now we show general formulae for the length of the latera recta of the hyperbola in terms of the coefficients of the general equation of the hyperbola, the conic discriminant, the conic radical, the conic discriminant and the conic determinant.

Corollary 15

(The general formula for the length of the latera recta of the hyperbola). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate hyperbola on the Cartesian plane, then the length of its two latera recta is given by the following general formula:

$${\ell }_r=\frac{4}{-\mathrm{sgn}(\Delta )(A+C)+\left|R\right|}\sqrt{\frac{2\left[\mathrm{sgn}(\Delta )(A+C)+\left|R\right|\right]|\Delta |}{d}}>0.$$

Proof. 

According to [Ref. [10], p. 211], the length of the latera recta of any non-degenerate hyperbola is given by ${\ell }_r={\displaystyle \frac{2b^2}{a}}>0$. Then, Corollary 14, Lemma 3(iii), and Proposition 5 imply

$${\ell }_r=\frac{2{\left(2\sqrt{\frac{2|\Delta |}{\left[-\mathrm{sgn}(\Delta )(A+C)+\left|R\right|\right]d}}\right)}^2}{2\sqrt{\frac{2|\Delta |}{\left[\mathrm{sgn}(\Delta )(A+C)+\left|R\right|\right]d}}}=\frac{4\sqrt{{4|\Delta |}^2}}{\left|-\mathrm{sgn}(\Delta )(A+C)+\left|R\right|\right|\sqrt{d^2}}\sqrt{\frac{\left[\mathrm{sgn}(\Delta )(A+C)+\left|R\right|\right]d}{2|\Delta |}}=\frac{4}{-\mathrm{sgn}(\Delta )(A+C)+\left|R\right|}\sqrt{\frac{2\left[\mathrm{sgn}(\Delta )(A+C)+\left|R\right|\right]|\Delta |}{d}}>0.$$

The proof is now complete. □

Finally, note that Corollary 14 also provides an alternative proof of Theorem 5(ii). Indeed, apply this result, along with the fact that

$$d=R^2-{(A+C)}^2$$

(22)

to the relation $\epsilon =c/a$.

6.2. Some Particularities of Equilateral Hyperbolas

This subsection is devoted to showing how the results presented in Section 6.1 are greatly simplified for the particular cases of equilateral hyperbolas.

Remark 11.

Consider that an equilateral hyperbola is defined as any non-degenerate hyperbola in which $a=b>0.$ So, according to [2], if (3) corresponds to a non-degenerate hyperbola, then it is equilateral if and only if $C=-A$.

Now we show formulae for the lengths of the three semiaxes of the equilateral hyperbola in terms of the coefficients of the general equation of the hyperbola, the conic discriminant, the conic radical, the conic discriminant and the conic determinant. Its formulae are direct consequences of Corollary 14, Remark 11 and the relation $a^2+b^2=c^2$, which holds for any non-degenerate hyperbola (see [Ref. [7], pp. 173–177]).

Corollary 16

(The formulae for the lengths of the three semiaxes of the equilateral hyperbola). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to an equilateral hyperbola on the Cartesian plane, then $a=b=2\sqrt{\frac{2|\Delta |}{\left|R\right|d}}$ and $c=4\sqrt{\frac{|\Delta |}{\left|R\right|d}}.$

The following is a useful simplification of the formula of the length of the latus rectum of the equilateral hyperbola. It is an immediate consequence of Corollaries 15 and 16, and Remark 11.

Corollary 17.

Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to an equilateral hyperbola on the Cartesian plane. Then, the length of its two latera recta is given by ${\ell }_r=2a=2b=\sqrt{2}c=4\sqrt{\frac{2|\Delta |}{\left|R\right|d}}>0.$

The following remark reveals a curious relationship between the conic radical and the conic discriminant that holds only for the equilateral hyperbolas.

Remark 12.

Observe that if (3) corresponds to an equilateral hyperbola, then (22) and Remark 11 imply $d=4A^2+B^2=B^2+4C^2=R^2>0$. This explains why $c={\displaystyle \frac{4\sqrt{|R\Delta |}}{d}}=4\sqrt{\frac{|\Delta |}{\left|R\right|d}}$ in this kind of hyperbolas, according to Corollaries 14 and 16.

6.3. The Equation of the Focal Axis of the Hyperbola

Corollaries 18 and 19 present general formulae for the focal axis of non-degenerate hyperbolas in terms of the coefficients of the general equation of the hyperbola, the conic discriminant, the conic radical, the conic discriminant and the conic determinant.

Corollary 18.

Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate hyperbola on the Cartesian Plane, then the equation of the straight line that corresponds to its focal axis is given as follows:

(i) 

If $\mathrm{sgn}\left(R\right)=\mathrm{sgn}(\Delta )\ne 0$, then the equation is

$$y=\sqrt{\frac{C-A+R}{A-C+R}}x+\frac{\sqrt{A-C+R}(2AE-BD)-\sqrt{C-A+R}(2CD-BE)}{\sqrt{A-C+R}d}.$$

(ii) 

If $\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )\ne 0$, then the equation is

$$y=-\sqrt{\frac{A-C+R}{C-A+R}}x+\frac{\sqrt{A-C+R}(2CD-BE)+\sqrt{C-A+R}(2AE-BD)}{\sqrt{C-A+R}d},wheneverB\ne 0;$$

otherwise, the equation is $x=-{\displaystyle \frac{D}{2A}}$.

Proof. 

(i) According to Remarks 1 and 2(b) and Theorem 21(i), the semiaxis a is on to the $x^{″}$-axis of Figure 1. Hence, Theorem 7(i) and Remark 5 yield the result in this case.

(ii) According to Remarks 1 and 2(b) and Theorem 21(ii), the semiaxis a is on the $y^{″}$-axis of Figure 1. Hence, Theorem 7(ii) and Remark 5 yield the result in this other case.

This completes the proof. □

The proof of the following result is omitted because it is similar to that of Corollary 18. The sole difference is that Corollary 4 is applied instead of Theorem 7, whereas (12) guarantees $\mathrm{sgn}\left(B\right)=\mathrm{sgn}\left(R\right)\ne 0$ whenever $\theta \ne 0\ne B$.

Corollary 19.

Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate hyperbola on the Cartesian plane, rotated by an angle $\theta \in \left(0,\frac{\pi }{2}\right)$. Then, the equation of the straight line that corresponds to its focal axis is given as follows:

(i) 

If $\mathrm{sgn}\left(R\right)=\mathrm{sgn}(\Delta )\ne 0$, then the equation is

$$y=\left(\frac{C-A+R}{B}\right)x+\frac{B(2AE-BD)-(C-A+R)(2CD-BE)}{Bd}.$$

(ii) 

If $\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )\ne 0$, then the equation is

$$y=\left(\frac{B}{A-C-R}\right)x+\frac{(A-C-R)(2AE-BD)-B(2CD-BE)}{(A-C-R)d}.$$

It is clear that Corollary 18 is more general than Corollary 19 because Corollary 18 works for all the non-degenerate hyperbolas, rotated or not. Meanwhile, Corollary 19 works only for the rotated hyperbolas.

6.4. On the Extreme and Focal Points of the Hyperbola

We devote this subsection to presenting all the general formulae for the coordinates of the extreme points (vertices and co-vertices) and the foci of any non-degenerate hyperbola. We start by stating a general theorem of the vertices of the hyperbola. Since its proof is analogous to the proof of Theorem 14; with the only difference that Theorem 21 is applied instead of Lemma 7, we omit it.

Theorem 22

(General theorem of the vertices of the hyperbola). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate hyperbola on the Cartesian plane, then its vertices are given as follows:

(i) 

If $\mathrm{sgn}\left(R\right)=\mathrm{sgn}(\Delta )\ne 0$, then

$${\mathcal{V}}_{1,2}\left(\frac{2CD-BE}{d}\pm 2\sqrt{\frac{(A-C+R)\Delta }{(A+C+R)Rd}},\frac{2AE-BD}{d}\pm 2\sqrt{\frac{(C-A+R)\Delta }{(A+C+R)Rd}}\right).$$

(ii) 

If $\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )\ne 0$, then

$${\mathcal{V}}_{1,2}\left(\frac{2CD-BE}{d}\mp 2\sqrt{\frac{(C-A+R)\Delta }{(A+C-R)Rd}},\frac{2AE-BD}{d}\pm 2\sqrt{\frac{(A-C+R)\Delta }{(A+C-R)Rd}}\right).$$

Now we present a general result for the co-vertices of the non-degenerate hyperbola in terms of the coefficients of the general equation of the hyperbola, the conic discriminant, the conic radical, the conic discriminant and the conic determinant.

Theorem 23

(General theorem of the co-vertices of the hyperbola). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate hyperbola on the Cartesian plane, then its co-vertices are given as follows:

(i) 

If $\mathrm{sgn}\left(R\right)=\mathrm{sgn}(\Delta )\ne 0$, then

$${\mathcal{B}}_{1,2}\left(\frac{2CD-BE}{d}\mp 2\sqrt{\frac{(A-C-R)\Delta }{(A+C-R)Rd}},\frac{2AE-BD}{d}\pm 2\sqrt{\frac{(A-C+R)\Delta }{(R-A-C)Rd}}\right).$$

(ii) 

If $\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )\ne 0$, then

$${\mathcal{B}}_{1,2}\left(\frac{2CD-BE}{d}\pm 2\sqrt{\frac{(C-A-R)\Delta }{(A+C+R)Rd}},\frac{2AE-BD}{d}\pm 2\sqrt{\frac{(A-C-R)\Delta }{(A+C+R)Rd}}\right).$$

Proof. 

This is analogous to the proof of Theorem 15, with the only difference that Theorem 21 is applied instead of Lemma 7. □

Now we present a general result for the foci of the non-degenerate hyperbola in terms of the coefficients of the general equation of the hyperbola, the conic discriminant, the conic radical, the conic discriminant and the conic determinant. Its proof is analogous to that of Theorem 16, with the only difference that Theorem 21 is applied instead of Lemma 7. Therefore, we leave the details to the interested reader.

Theorem 24

(General theorem of the foci of the hyperbola). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate hyperbola on the Cartesian plane. Then, its focal points are given as follows:

(i) 

If $\mathrm{sgn}\left(R\right)=\mathrm{sgn}(\Delta )\ne 0$, then

$${\mathcal{F}}_{1,2}\left(\frac{2CD-BE\pm 2\sqrt{2(A-C+R)\Delta }}{d},\frac{2AE-BD\pm 2\sqrt{2(C-A+R)\Delta }}{d}\right).$$

(ii) 

If $\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )\ne 0$, then

$${\mathcal{F}}_{1,2}\left(\frac{2CD-BE\mp 2\sqrt{2(A-C-R)\Delta }}{d},\frac{2AE-BD\pm 2\sqrt{2(C-A-R)\Delta }}{d}\right).$$

6.5. The Equations of the Directrices and the Latera Recta of the Hyperbola

The following result states a simple formula for the distance of the center of a non-degenerate hyperbola, to each of its directrices.

Lemma 9.

Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate hyperbola on the Cartesian plane, then the least distance between the central point of this hyperbola and any of its two directrices is given by the following general formula:

$${\delta }_D=\frac{2}{\mathrm{sgn}(\Delta )(A+C)+\left|R\right|}\sqrt{\left|\frac{\Delta }{R}\right|}>0.$$

Proof. 

By Corollary 14, Remark 9, and Proposition 5, we have that

$$\begin{array}{c}\hfill {\delta }_D=\frac{a^2}{c}=\frac{{\left(2\sqrt{\frac{2|\Delta |}{\left[\mathrm{sgn}(\Delta )(A+C)+\left|R\right|\right]d}}\right)}^2}{\left(\frac{4\sqrt{|R\Delta |}}{d}\right)}=\frac{2\sqrt{{|\Delta |}^2}d}{\left|\mathrm{sgn}(\Delta )(A+C)+\left|R\right|\right|\sqrt{\left|R\right||\Delta |}d}=\frac{2}{\mathrm{sgn}(\Delta )(A+C)+\left|R\right|}\sqrt{\left|\frac{\Delta }{R}\right|}>0.\end{array}$$

This proves the result. □

We now point out an important relation between the coefficients of the general equation of the hyperbola and the conic determinant.

Proposition 7.

Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $\mathrm{sgn}(A-C)=\mathrm{sgn}(\Delta )\ne 0$, then $\mathrm{sgn}(\Delta )(A+C)+|A-C|=2A\mathrm{sgn}(\Delta ).$

Proof. 

Case 1. If $\mathrm{sgn}(A-C)=\mathrm{sgn}(\Delta )=1$, then $|A-C|=A-C>0$. So, $\mathrm{sgn}(\Delta )(A+C)+|A-C|=\left(1\right)(A+C)+(A-C)=2A=2A\mathrm{sgn}(\Delta ).$

Case 2. If $\mathrm{sgn}(A-C)=\mathrm{sgn}(\Delta )=-1$, then $|A-C|=C-A>0$. So, $\mathrm{sgn}(\Delta )(A+C)+|A-C|=(-1)(A+C)+(C-A)=-2A=2A\mathrm{sgn}(\Delta )$. □

Now we present a general characterization of the equations of the directrices of the non-degenerate hyperbola in terms of the coefficients of the general equation of the hyperbola, the conic discriminant, the conic radical, the conic discriminant and the conic determinant.

Theorem 25

(General characterization of the equations of the directrices of hyperbolas). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate hyperbola on the Cartesian plane. Then, the equations of the straight lines that correspond to its directrices are given as follows:

(i) 

If $\mathrm{sgn}\left(R\right)=\mathrm{sgn}(\Delta )\ne 0$, then the equations are

$$y=-\sqrt{\frac{A-C+R}{C-A+R}}x+\left[\frac{\sqrt{A-C+R}(2CD-BE)+\sqrt{C-A+R}(2AE-BD)}{\sqrt{C-A+R}d}\pm \frac{2}{\mathrm{sgn}(\Delta )(A+C)+\left|R\right|}\sqrt{\frac{2\Delta }{C-A+R}}\right],wheneverB\ne 0;$$

otherwise, the equations are $x=-\frac{1}{2A}\left(D\pm \sqrt{\frac{4ACF-A{E}^2-C{D}^2}{A-C}}\right)$.

(ii) 

If $\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )\ne 0$, then the equations are

$$y=\sqrt{\frac{C-A+R}{A-C+R}}x+\left[\frac{\sqrt{A-C+R}(2AE-BD)-\sqrt{C-A+R}(2CD-BE)}{\sqrt{A-C+R}d}\pm \frac{2}{\mathrm{sgn}(\Delta )(A+C)+\left|R\right|}\sqrt{\frac{2\Delta }{C-A-R}}\right].$$

Proof. 

(i) Corollary 18(i) implies that the two directrices of the hyperbola are parallel to the $y^{″}$-axis of Figure 1, which is also in the middle of both directrices. Now consider the following two cases:

Case 1. Suppose that $\theta \in \left(0,\frac{\pi }{2}\right)$, then $B\ne 0$. Hence, Corollary 6, Remarks 4(ii) and 5, Proposition 2(ii) and Lemma 9 imply that the equations of the directrices are given by $y=mx+\left(b\pm {\delta }_D\sqrt{m^2+1}\right)$, where

$$\begin{array}{ccc}\hfill m& =& -\sqrt{\frac{A-C+R}{C-A+R}},b=\frac{\sqrt{A-C+R}(2CD-BE)+\sqrt{C-A+R}(2AE-BD)}{\sqrt{C-A+R}d}\mathrm{and}\hfill \\ \hfill {\delta }_D\sqrt{m^2+1}& =& \left[\frac{2}{\mathrm{sgn}(\Delta )(A+C)+\left|R\right|}\sqrt{\left|\frac{\Delta }{R}\right|}\right]\sqrt{\frac{2R}{C-A+R}}=\frac{2}{\mathrm{sgn}(\Delta )(A+C)+\left|R\right|}\sqrt{\left(\frac{\Delta }{R}\right)\left(\frac{2R}{C-A+R}\right)}=\frac{2}{\mathrm{sgn}(\Delta )(A+C)+\left|R\right|}\sqrt{\frac{2\Delta }{C-A+R}}.\hfill \end{array}$$

Case 2. Suppose that $\theta =B=0$; then, (10) guarantees $R=A-C$, whereas the equations of the two directrices are given by $x=h\pm \frac{a}{\epsilon }$ (see ref. [6], p. 117). Hence, (5), (8), Remark 9, and Proposition 7 imply

$$\begin{array}{ccc}\hfill x& =& h\pm \frac{a}{\epsilon }=\frac{2CD-BE}{B^2-4AC}\pm \frac{a^2}{c}=\frac{2CD-\left(0\right)E}{0^2-4AC}\pm {\delta }_D=-\frac{D}{2A}\pm \frac{2}{\mathrm{sgn}(\Delta )(A+C)+\left|R\right|}\sqrt{\frac{\Delta }{R}}\hfill \\ & =& -\frac{D}{2A}\pm \frac{2}{2A\mathrm{sgn}(\Delta )}\sqrt{\frac{1}{A-C}\left[\frac{\left(0\right)DE-A{E}^2-C{D}^2-(0^2-4AC)F}{4}\right]}=-\frac{D}{2A}\pm \frac{1}{2A\mathrm{sgn}(\Delta )}\sqrt{\frac{4ACF-A{E}^2-C{D}^2}{A-C}}.\hfill \end{array}$$

Finally, observe that $\mathrm{sgn}(\Delta )\ne 0$ becomes redundant in this expression due to the double sign, so this yields the result.

(ii) Corollary 18(ii) implies that the two directrices of the hyperbola are parallel to the $x^{″}$–axis of Figure 1. Hence, Corollary 6, Remarks 4(i) and 5, Proposition 2(i) and Lemma 9 imply that the equations of the directrices are $y=mx+\left(b\pm {\delta }_D\sqrt{m^2+1}\right)$, where

$$\begin{array}{ccc}\hfill m& =& \sqrt{\frac{C-A+R}{A-C+R}},b=\frac{\sqrt{A-C+R}(2AE-BD)-\sqrt{C-A+R}(2CD-BE)}{\sqrt{A-C+Rd}}\mathrm{and}\hfill \\ \hfill {\delta }_D\sqrt{m^2+1}& =& \left[\frac{2}{\mathrm{sgn}(\Delta )(A+C)+\left|R\right|}\sqrt{\left|\frac{\Delta }{R}\right|}\right]\sqrt{\frac{2R}{A-C+R}}=\frac{2}{\mathrm{sgn}(\Delta )(A+C)+\left|R\right|}\sqrt{\left(-\frac{\Delta }{R}\right)\left(\frac{2R}{A-C+R}\right)}=\frac{2}{\mathrm{sgn}(\Delta )(A+C)+\left|R\right|}\sqrt{\frac{2\Delta }{C-A-R}}.\hfill \end{array}$$

This proves the result. □

Now we present a characterization of the equations of the directrices of the non-degenerate rotated hyperbola in terms of the coefficients of the general equation of the hyperbola, the conic discriminant, the conic radical, the conic discriminant and the conic determinant. Its proof is immediate and thus is omitted, as it is similar to the proofs of case 1 of item (i) and item (ii) of Theorem 25. The only difference is that Corollary 19 is used instead of Corollary 18.

Theorem 26

(Characterization of the equations of the directrices of rotated hyperbolas). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate hyperbola on the Cartesian plane, rotated by an angle $\theta \in (0,\frac{\pi }{2})$. Then, the equations of the straight lines that correspond to its directrices are given as follows:

(i) 

If $\mathrm{sgn}\left(R\right)=\mathrm{sgn}(\Delta )\ne 0$, then the equations are

$$\begin{array}{c}\hfill y=\left(\frac{B}{A-C-R}\right)x+\left[\frac{(A-C-R)(2AE-BD)-B(2CD-BE)}{(A-C-R)d}\pm \frac{2}{\mathrm{sgn}(\Delta )(A+C)+\left|R\right|}\sqrt{\frac{2\Delta }{C-A+R}}\right].\end{array}$$

(ii) 

If $\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )\ne 0$, then the equations are

$$\begin{array}{c}\hfill y=\left(\frac{C-A+R}{B}\right)x+\left[\frac{B(2AE-BD)-(C-A+R)(2CD-BE)}{Bd}\pm \frac{2}{\mathrm{sgn}(\Delta )(A+C)+\left|R\right|}\sqrt{\frac{2\Delta }{C-A-R}}\right].\end{array}$$

Now we present a general characterization of the equations of latera recta of the non-degenerate hyperbola in terms of the coefficients of the general equation of the hyperbola, the conic discriminant, the conic radical, the conic discriminant and the conic determinant. Its proof is straightforward and is thus omitted, as it is similar to the proof of Theorem 25. The only difference is that the formulae given for c in Theorem 21 are applied instead of Lemma 9.

Theorem 27

(General characterization of the equations of latera recta of hyperbolas). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate hyperbola on the Cartesian plane. Then, the equations of the straight lines that correspond to its latera recta are given as follows:

(i) 

If $\mathrm{sgn}\left(R\right)=\mathrm{sgn}(\Delta )\ne 0$, then the equations are

$$\begin{array}{c}\hfill y=-\sqrt{\frac{A-C+R}{C-A+R}}x+\frac{1}{d}\left[\frac{\sqrt{A-C+R}(2CD-BE)+\sqrt{C-A+R}(2AE-BD)}{\sqrt{C-A+R}}\pm 4R\sqrt{\frac{2\Delta }{C-A+R}}\right],wheneverB\ne 0;\end{array}$$

otherwise, the equations are $x=-{\displaystyle \frac{1}{2A}}\left(D\pm \sqrt{{\displaystyle \frac{(A-C)(4ACF-A{E}^2-C{D}^2)}{C}}}\right)$.

(ii) 

If $\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )\ne 0$, then the equations are

$$\begin{array}{c}\hfill y=\sqrt{\frac{C-A+R}{A-C+R}}x+\frac{1}{d}\left[\frac{\sqrt{A-C+R}(2AE-BD)-\sqrt{C-A+R}(2CD-BE)}{\sqrt{A-C+R}}\pm 4R\sqrt{\frac{2\Delta }{C-A-R}}\right].\end{array}$$

Now we present a general characterization of the equations of latera recta of the non-degenerate rotated hyperbola in terms of the coefficients of the general equation of the hyperbola, the conic discriminant, the conic radical, the conic discriminant and the conic determinant. Its proof is straightforward, as it is similar to the proofs of case 1 of items (i) and (ii) of Theorem 27. The only difference is that Corollary 19 is used instead of Corollary 18.

Theorem 28

(Characterization of the equations of the latera recta of rotated hyperbolas). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate hyperbola on the Cartesian plane, rotated by an angle $\theta \in (0,\frac{\pi }{2})$. Then, the equations of the straight lines that correspond to its latera recta are given as follows:

(i) 

If $\mathrm{sgn}\left(R\right)=\mathrm{sgn}(\Delta )\ne 0$, then the equations are

$$\begin{array}{c}\hfill y=\left(\frac{B}{A-C-R}\right)x+\frac{1}{d}\left[\frac{(A-C-R)(2AE-BD)-B(2CD-BE)}{A-C-R}\pm 4R\sqrt{\frac{2\Delta }{C-A+R}}\right].\end{array}$$

(ii) 

If $\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )\ne 0$, then equations are

$$\begin{array}{c}\hfill y=\left(\frac{C-A+R}{B}\right)x+\frac{1}{d}\left[\frac{B(2AE-BD)-(C-A+R)(2CD-BE)}{B}\pm 4R\sqrt{\frac{2\Delta }{C-A-R}}\right].\end{array}$$

It is clear that Theorem 25 is more general than Theorem 26, in the same sense of Remark 4, with respect to Theorem 6. Meanwhile, Theorem 27 is more general than Theorem 28 in the very same sense.

6.6. The Equations of the Asymptotes of the Hyperbola

Now we present a general characterization of the equations of the asymptotes of the non-degenerate hyperbola in terms of the coefficients of the general equation of the hyperbola, the conic discriminant, the conic radical, the conic discriminant and the conic determinant.

Theorem 29

(General characterization of the equations of the asymptotes of the hyperbola). Let $A,B,C,D,E,F\in \mathbb{R}$ be such that $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ corresponds to a non-degenerate hyperbola rotated by some angle $\theta \in [0,\pi /2)$. Then, the equations of its two asymptotes are given as follows:

(i) 

$y=\left({\displaystyle \frac{-B\pm \sqrt{d}}{2C}}\right)x+{\displaystyle \frac{2C(2AE-BD)+(2CD-BE)(B\mp \sqrt{d})}{2Cd}}$, whether $C\ne 0$;

(ii) 

$y=-{\displaystyle \frac{A}{B}}x+{\displaystyle \frac{AE-BD}{d}}$ and $x=-{\displaystyle \frac{E}{B}}$ with $B\ne 0$, whether $C=0$.

Proof. 

(i) Suppose that $C\ne 0$. If (3) corresponds to a non-degenerate hyperbola and the equations of its two asymptotes are $y=m_{1}x+b_1$ and $y=m_{2}x+b_2$, then the equation that represents these asymptotes at the same time is given as follows (see [2]):

$$\begin{array}{cc}\hfill & C(m_{1}x-y+b_1)(m_{2}x-y+b_2)=C{m}_1{m}_2{x}^2-C(m_1+m_2)xy\hfill \\ & +C{y}^2+C(b_1{m}_2+b_2{m}_1)x-C(b_1+b_2)y+C{b}_1{b}_2=0.\hfill \end{array}$$

(23)

Thus, (23) gives the relations $A=C{m}_1{m}_2$ and $B=-C(m_1+m_2)$, which imply the following non-linear system of simultaneous equations:

$$m_1+m_2=-\frac{B}{C}\mathrm{whereas}{m}_1{m}_2=\frac{A}{C}.$$

(24)

Hence, (24) is equivalent to the quadratic equation $C{m}^2+Bm+A=0$, whose two roots are the solutions of (24). So, the quadratic formula implies $m={\displaystyle \frac{-B\pm \sqrt{B^2-4AC}}{2C}}={\displaystyle \frac{-B\pm \sqrt{d}}{2C}}$. That is, $m_1={\displaystyle \frac{-B+\sqrt{d}}{2C}}$ and $m_2={\displaystyle \frac{-B-\sqrt{d}}{2C}}.$ (Note that Corollary 2(iii) guarantees $m_1,m_2\in \mathbb{R}$ with $m_1\ne m_2$). On the other hand, if $\mathcal{C}(h,k)$ is the center of the hyperbola in question, then the relations $k=m_{1}h+b_1$ and $k=m_{2}h+b_2$ hold because both asymptotes intersect at $\mathcal{C}$. Therefore, (8) and (18) imply

$$\begin{array}{ccc}\hfill b_1& =& k-m_{1}h=\frac{2AE-BD}{d}-\left(\frac{-B+\sqrt{d}}{2C}\right)\left(\frac{2CD-BE}{d}\right)=\frac{2C(2AE-BD)+(2CD-BE)(B-\sqrt{d})}{2Cd},\hfill \\ \hfill b_2& =& k-m_{2}h=\frac{2AE-BD}{d}-\left(\frac{-B-\sqrt{d}}{2C}\right)\left(\frac{2CD-BE}{d}\right)=\frac{2C(2AE-BD)+(2CD-BE)(B+\sqrt{d})}{2Cd}.\hfill \end{array}$$

(ii) Suppose that $C=0$. Then (18) and Corollary 2(iii) imply $d=B^2>0$, which guarantees $B\ne 0$. In this case, the equations of the two asymptotes of the hyperbola in question are $y=mx+b$ and $x=h$. So the equation that represents these asymptotes at the same time is given as follows (see [2]):

$$-B(mx-y+b)(x-h)=-Bm{x}^2+Bxy+B(mh-b)x-Bhy+Bbh=0.$$

(25)

So, (25) gives the relations $A=-Bm$, $D=B(mh-b)$ and $E=-Bh$. Hence, $m=-{\displaystyle \frac{A}{B}}$, $h=-{\displaystyle \frac{E}{B}}$, $b=mh-{\displaystyle \frac{D}{B}}=\left(-{\displaystyle \frac{A}{B}}\right)\left(-{\displaystyle \frac{E}{B}}\right)-{\displaystyle \frac{D}{B}}={\displaystyle \frac{AE-BD}{B^2}}={\displaystyle \frac{AE-BD}{d}},$ which are well-defined due to the fact that $B\ne 0$.

This completes the proof. □

Now we present a general characterization of the equations of the asymptotes of the non-degenerate non-rotated hyperbola in terms of the coefficients of the general equation of the hyperbola, the conic discriminant, the conic radical, the conic discriminant and the conic determinant.

Corollary 20

(Characterization of the equations of the asymptotes of non-rotated hyperbolas). Let $A,C,D,E,F\in \mathbb{R}$ be such that the equation $A{x}^2+C{y}^2+Dx+Ey+F=0$ corresponds to a non-rotated hyperbola. Then the equations of its asymptotes are given by

$$y=\left(\pm \frac{\sqrt{-AC}}{C}\right)x-\frac{AE\mp D\sqrt{-AC}}{2AC}.$$

Proof. 

First of all, note that if $B=0$, then (18) and Corollary 2(iii) imply $d=-4AC>0$, which guarantees $A\ne 0\ne C$. Hence, Theorem 29(i) implies that the equations of the two asymptotes of the hyperbola in question are given as follows:

$$\begin{array}{c}\hfill y=\left(\frac{0\pm \sqrt{d}}{2C}\right)x+\frac{2C\left(2AE-0\right)+\left(2CD-0\right)\left(0\mp \sqrt{d}\right)}{2Cd}=\left(\pm \frac{\sqrt{-4AC}}{2C}\right)x+\frac{2C\left(2AE\mp D\sqrt{-4AC}\right)}{2C\left(-4AC\right)}=\left(\pm \frac{\sqrt{-AC}}{C}\right)x-\frac{AE\mp D\sqrt{-AC}}{2AC}.\end{array}$$

This completes the proof. □

6.7. Examples of Non-Degenerate Hyperbolas

We complete this section by giving the following illustrations.

Example 6.

Suppose that $5x^2+12xy-13x-24y+5=0.$ Then $R=\mathrm{sgn}\left(12\right)\sqrt{{\left(5-0\right)}^2+{12}^2}=13$ and $A=5\ne -C=0$, thus $\left|R\right|=13>\left|A+C\right|=\left|5+0\right|=5$. In addition, (5) implies $\Delta =36$, so $\mathrm{sgn}\left(R\right)=\mathrm{sgn}\left(\Delta \right)=1$. Thereby, Lemma 3(iii) and Remark 11 guarantee that the given equation corresponds to a non-equilateral hyperbola, whose eccentricity is given by Theorem 5(ii) as $\epsilon =\sqrt{{\displaystyle \frac{2\left|13\right|}{-\mathrm{sgn}\left(36\right)(5+0)+\left|13\right|}}}={\displaystyle \frac{\sqrt{13}}{2}}.$ Now consider that (18) gives $d=144$, thus Corollary 15 implies

$${\ell }_r=\frac{4}{-\left(1\right)\left(5+0\right)+\left|13\right|}\sqrt{\frac{2\left[\left(1\right)\left(5+0\right)+\left|13\right|\right]\left|36\right|}{144}}=\frac{3}{2}.$$

Whereas Corollary 14 yields

$$a=2\sqrt{\frac{2\left|36\right|}{\left[\left(1\right)\left(5+0\right)+\left|13\right|\right]\left(144\right)}}=\frac{1}{3},b=2\sqrt{\frac{2\left|36\right|}{\left[-\left(1\right)\left(5+0\right)+\left|13\right|\right]\left(144\right)}}=\frac{1}{2}andc=\frac{4\sqrt{\left|\left(13\right)\left(36\right)\right|}}{144}=\frac{\sqrt{13}}{6}.$$

Additionally, (8) implies $h=2$ and $k=-\frac{7}{12}.$ So, the center of the hyperbola is $\mathcal{C}\left(2,-\frac{7}{12}\right)$. Hence, Theorem 22(i) implies that the two vertices of the hyperbola are given by

$${\mathcal{V}}_{1,2}\left(2\pm 2\sqrt{\frac{(5-0+13)\left(36\right)}{(5+0+13)\left(13\right)\left(144\right)}},-\frac{7}{12}\pm 2\sqrt{\frac{(0-5+13)\left(36\right)}{(5+0+13)\left(13\right)\left(144\right)}}\right).$$

Thus ${\mathcal{V}}_1\left(\frac{26+\sqrt{13}}{13},\frac{-91+8\sqrt{13}}{156}\right)$ and ${\mathcal{V}}_2\left(\frac{26-\sqrt{13}}{13},\frac{-91-8\sqrt{13}}{156}\right).$ Meanwhile, Theorem 23(i) implies that the two co-vertices of the hyperbola are given by

$${\mathcal{B}}_{1,2}\left(2\mp 2\sqrt{\frac{(5-0-13)\left(36\right)}{(5+0-13)\left(13\right)\left(144\right)}},-\frac{7}{12}\pm 2\sqrt{\frac{(5-0+13)\left(36\right)}{(13-5-0)\left(13\right)\left(144\right)}}\right).$$

So ${\mathcal{B}}_1\left(\frac{26-\sqrt{13}}{13},\frac{-91+18\sqrt{13}}{156}\right)$ and ${\mathcal{B}}_2\left(\frac{26+\sqrt{13}}{13},\frac{-91-18\sqrt{13}}{156}\right)$, and Theorem 24(i) implies that the foci of the hyperbola are given by

$${\mathcal{F}}_{1,2}\left(2\pm \frac{2\sqrt{2(5-0+13)\left(36\right)}}{144},-\frac{7}{12}\pm \frac{2\sqrt{2(0-5+13)\left(36\right)}}{144}\right).$$

Hence ${\mathcal{F}}_1\left(\frac{5}{2},-\frac{1}{4}\right)$ and ${\mathcal{F}}_2\left(\frac{3}{2},-\frac{11}{12}\right).$ On the other hand, Corollary 18(i) guarantees that the focal axis of the hyperbola is given by

$$\begin{array}{c}\hfill y=\sqrt{\frac{0-5+13}{5-0+13}}x+\frac{\sqrt{5-0+13}\left[2\left(5\right)(-24)-\left(12\right)(-13)\right]-\sqrt{0-5+13}\left[2\left(0\right)(-13)-\left(12\right)(-24)\right]}{\sqrt{5-0+13}\left(144\right)},\end{array}$$

which is reduced to $y=\frac{2}{3}x-\frac{23}{12}$. In addition, Theorem 25(i) implies that the equations of the two directrices of the hyperbola are given by

$$\begin{array}{c}\hfill y=-\sqrt{\frac{5-0+13}{0-5+13}}x+\left[\frac{\sqrt{5-0+13}[2\left(0\right)(-13)-\left(12\right)(-24)]+\sqrt{0-5+13}[2\left(5\right)(-24)-\left(12\right)(-13)]}{\sqrt{0-5+13}\left(144\right)}\pm \frac{2}{\left(1\right)(5+0)+\left|13\right|}\sqrt{\frac{2\left(36\right)}{0-5+13}}\right],\end{array}$$

which is reduced to $y=-\frac{3}{2}x+\frac{11}{4}$ and $y=-\frac{3}{2}x+\frac{25}{12}$. Meanwhile, Theorem 27(i) implies that the equations of the latera recta of the hyperbola are given by

$$\begin{array}{c}\hfill y=-\sqrt{\frac{5-0+13}{0-5+13}}x+\frac{1}{144}\left[\frac{\sqrt{5-0+13}[2\left(0\right)(-13)-\left(12\right)(-24)]+\sqrt{0-5+13}[2\left(5\right)(-24)-\left(12\right)(-13)]}{\sqrt{0-5+13}}\pm 4\left(13\right)\sqrt{\frac{2\left(36\right)}{0-5+13}}\right],\end{array}$$

which is reduced to $y=-\frac{3}{2}x+\frac{7}{2}$ and $y=-\frac{3}{2}x+\frac{4}{3}$. Note that all the previous equations can also be respectively obtained by Theorems 26(i) and 28(i). So, Corollary 5 implies that the general equations of the two degenerate cases of parabolas that respectively correspond to the directrices and latera recta of the hyperbola are $108x^2+144xy+48y^2-348x-232y+275=0$ and $27x^2+36xy+12y^2-87x-58y+56=0.$ Moreover, Theorem 29(ii) implies that the equations of the two asymptotes of the hyperbola are respectively given by $y=-\frac{5}{12}x+\frac{\left(5\right)(-24)-\left(12\right)(-13)}{144}$ and $x=-\frac{(-24)}{12}$. That is, $y=-\frac{5}{12}x+\frac{1}{4}$ and $x=2$.

Finally, note that the rotation angle was completely unneeded to obtain all these results. However, if it is required, then it can be obtained by applying Corollary 3 as $tan\theta =\sqrt{{\displaystyle \frac{0-5+13}{5-0+13}}}={\displaystyle \frac{2}{3}}$, thus $\theta =arctan\left({\displaystyle \frac{2}{3}}\right)\approx 33.69°$. See Figure 10.

Example 7.

Suppose that $2x^2-3xy-2y^2+13x+9y-23=0.$ Then

$$R=\mathrm{sgn}\left(-3\right)\sqrt{{\left[2-\left(-2\right)\right]}^2+{\left(-3\right)}^2}=-5,$$

Now observe that $A=-C=2$; thus, $\left|R\right|=5>\left|A+C\right|=\left|2+\left(-2\right)\right|=0.$ In addition, (5) implies $\Delta =100$, so $\mathrm{sgn}\left(R\right)=-\mathrm{sgn}\left(\Delta \right)=-1$. Thereby, Lemma 3(iii) and Remark 11 guarantee that the given equation corresponds to a non-degenerate equilateral hyperbola, whose eccentricity is given by Theorem 5(i) as $\epsilon =\sqrt{{\displaystyle \frac{2\left|-5\right|}{\left|2+\left(-2\right)\right|+\left|-5\right|}}}=\sqrt{2}.$ Now consider that (18) gives $d=25$, thus Corollary 17 implies ${\ell }_r=4\sqrt{\frac{2\left|100\right|}{|-5|\left(25\right)}}={\displaystyle \frac{8\sqrt{10}}{5}}$, whereas Corollary 16 implies $a=b=2\sqrt{\frac{2\left|100\right|}{|-5|\left(25\right)}}=\frac{4\sqrt{10}}{5}$ and $c=4\sqrt{\frac{\left|100\right|}{|-5|\left(25\right)}}=\frac{8}{\sqrt{5}}$. Additionally, (8) implies $h=-1$ and $k=3$. So, the center of the hyperbola is $\mathcal{C}(-1,3)$. Hence, Theorem 22(ii) implies that the two vertices of the hyperbola are given by

$${\mathcal{V}}_{1,2}\left(-1\mp 2\sqrt{\frac{[-2-2+(-5\left)\right]\left(100\right)}{[2+(-2)-(-5\left)\right](-5)\left(25\right)}},3\pm 2\sqrt{\frac{[2-(-2)+(-5\left)\right]\left(100\right)}{[2+(-2)-(-5\left)\right](-5)\left(25\right)}}\right).$$

Thus ${\mathcal{V}}_1\left(-\frac{17}{5},\frac{19}{5}\right)$ and ${\mathcal{V}}_2\left(\frac{7}{5},\frac{11}{5}\right)$. Meanwhile, Theorem 23(ii) implies that the two co-vertices of the hyperbola are given by

$${\mathcal{B}}_{1,2}\left(-1\pm 2\sqrt{\frac{[-2-2-(-5\left)\right]\left(100\right)}{[2+(-2)+(-5\left)\right](-5)\left(25\right)}},3\pm 2\sqrt{\frac{[2-(-2)-(-5\left)\right]\left(100\right)}{[2+(-2)+(-5\left)\right](-5)\left(25\right)}}\right).$$

So ${\mathcal{B}}_1\left(-\frac{1}{5},\frac{27}{5}\right)$ and ${\mathcal{B}}_2\left(-\frac{9}{5},\frac{3}{5}\right)$, and Theorem 24(ii) implies that the two foci of the hyperbola are given by

$${\mathcal{F}}_{1,2}\left(-1\mp \frac{2\sqrt{2[2-(-2)-(-5\left)\right]\left(100\right)}}{25},3\pm \frac{2\sqrt{2[-2-2-(-5\left)\right]\left(100\right)}}{25}\right),$$

Hence ${\mathcal{F}}_1\left(\frac{-5-12\sqrt{2}}{5},\frac{15+4\sqrt{2}}{5}\right)$ and ${\mathcal{F}}_2\left(\frac{-5+12\sqrt{2}}{5},\frac{15-4\sqrt{2}}{5}\right).$ On the other hand, Corollary 19(ii) guarantees that the equation of the focal axis of the hyperbola is given by

$$y=\left[\frac{(-3)}{2-(-2)-(-5)}\right]x+\frac{[2-(-2)-(-5\left)\right]\left[2\right(2\left)\right(9)-(-3\left)\right(13\left)\right]-(-3)\left[2\right(-2\left)\right(13)-(-3\left)\right(9\left)\right]}{[2-(-2)-(-5\left)\right]\left(25\right)},$$

which is reduced to $y=-{\displaystyle \frac{1}{3}}x+{\displaystyle \frac{8}{3}}$. In addition, Theorems 26(ii) and 28(ii) imply that the equations of the two directrices and the latera recta of the hyperbola are respectively given by

$$y=\left[\frac{-2-2+(-5)}{(-3)}\right]x+\left[\frac{(-3)\left(2\left(2\right)\left(9\right)-(-3)\left(13\right)\right)-\left(-2-2+(-5)\right)\left(2(-2)\left(13\right)-(-3)\left(9\right)\right)}{(-3)\left(25\right)}\pm \frac{2}{\left(1\right)\left(2+(-2)\right)+|-5|}\sqrt{\frac{2\left(100\right)}{-2-2-(-5)}}\right],$$

$$y=\left[\frac{-2-2+(-5)}{(-3)}\right]x+\frac{1}{25}\left[\frac{(-3)\left(2\left(2\right)\left(9\right)-(-3)\left(13\right)\right)-\left(-2-2+(-5)\right)\left(2(-2)\left(13\right)-(-3)\left(9\right)\right)}{(-3)}\pm 4(-5)\sqrt{\frac{2\left(100\right)}{-2-2-(-5)}}\right],$$

which are respectively reduced to $y=3x+(6\pm 4\sqrt{2})$, and $y=3x+(6\mp 8\sqrt{2})$. Note that these equations can also be respectively obtained by Theorems 25(ii) and 27(ii). So, Corollary 5 implies that the general equations of the two degenerate cases of parabolas that respectively correspond to the directrices and the latera recta of the hyperbola in question are $9x^2-6xy+y^2+36x-12y+4=0$ and $9x^2-6xy+y^2+36x-12y-92=0$. Moreover, Theorem 29(i) implies that the equations of the two asymptotes of this hyperbola are given by

$$y=\left[\frac{-(-3)\pm \sqrt{25}}{2(-2)}\right]x+\frac{2(-2)\left[2\left(2\right)\left(9\right)-(-3)\left(13\right)\right]+\left[2(-2)\left(13\right)-(-3)\left(9\right)\right]\left(-3\mp \sqrt{25}\right)}{2(-2)\left(25\right)},$$

which are reduced to $y=-2x+1$ and $y=\frac{1}{2}x+\frac{7}{2}$. Finally, note that the rotation angle was completely unneeded to obtain all these results. However, if it is required, then it can be obtained by applying Corollary 3 as $tan\theta =\sqrt{{\displaystyle \frac{-2-2+(-5)}{2-(-2)+(-5)}}}=3$, thus $\theta =arctan\left(3\right)\approx 71.57°$. See Figure 11.

Example 8.

Suppose that $2x^2-y^2-6y-3=0$. Then $R=2-(-1)=3$ and $A=2\ne -C=1$, thus $\left|R\right|=3>|A+C|=|2+(-1\left)\right|=1$. In addition, $\Delta =-12$, so $\mathrm{sgn}\left(R\right)=-\mathrm{sgn}(\Delta )=1$. Thereby, Lemma 3(iii) and Remark 11 guarantee that the given equation corresponds to a non-equilateral hyperbola, whose eccentricity is given by Theorem 5(ii) as $\epsilon =\sqrt{{\displaystyle \frac{2\left|3\right|}{-\mathrm{sgn}(-12)[2+(-1\left)\right]+\left|3\right|}}}=\sqrt{{\displaystyle \frac{3}{2}}}$. Now consider that (18) gives $d=8$, thus Corollary 15 implies ${\ell }_r={\displaystyle \frac{4}{-(-1)[2+(-1\left)\right]+\left|3\right|}}\sqrt{{\displaystyle \frac{2\left[\right(-1\left)\right(2+(-1))+|3\left|\right]|-12|}{8}}}=\sqrt{6}$. Whereas Corollary 14 implies

$$a=2\sqrt{\frac{2|-12|}{\left[(-1)\left(2+(-1)\right)+\left|3\right|\right]\left(8\right)}}=\sqrt{6},b=2\sqrt{\frac{2|-12|}{\left[-(-1)\left(2+(-1)\right)+\left|3\right|\right]\left(8\right)}}=\sqrt{3}andc=\frac{4\sqrt{\left|\right(3\left)\right(-12\left)\right|}}{8}=3.$$

Additionally, (8) implies $h=0$ and $k=-3$. So, the center of the hyperbola is $\mathcal{C}(0,-3)$. Hence, Theorem 22(ii) implies that the two vertices of the hyperbola are given by

$${\mathcal{V}}_{1,2}\left(0\mp 2\sqrt{\frac{(-1-2+3)(-12)}{(2+(-1)-3)\left(3\right)\left(8\right)}},-3\pm 2\sqrt{\frac{(2-(-1)+3)(-12)}{(2+(-1)-3)\left(3\right)\left(8\right)}}\right).$$

Thus ${\mathcal{V}}_1(0,-3+\sqrt{6})$ and ${\mathcal{V}}_2(0,-3-\sqrt{6}).$ Meanwhile, Theorem 23(ii) implies that the two co-vertices of the hyperbola are given by

$${\mathcal{B}}_{1,2}\left(0\pm 2\sqrt{\frac{(-1-2-3)(-12)}{(2+(-1)+3)\left(3\right)\left(8\right)}},-3\pm 2\sqrt{\frac{(2-(-1)-3)(-12)}{(2+(-1)+3)\left(3\right)\left(8\right)}}\right).$$

So ${\mathcal{B}}_1(\sqrt{3},-3)$ and ${\mathcal{B}}_2(-\sqrt{3},-3)$, and Theorem 24(ii) implies that the two foci of the hyperbola are given by

$${\mathcal{F}}_{1,2}\left(0\mp \frac{2\sqrt{2[2-(-1)-3](-12)}}{8},-3\pm \frac{2\sqrt{2(-1-2-3)(-12)}}{8}\right).$$

Hence ${\mathcal{F}}_1(0,0)$ and ${\mathcal{F}}_2(0,-6)$. On the other hand, Corollary 18(ii) guarantees that the equation of the focal axis of the hyperbola is given by $x=-{\displaystyle \frac{0}{2\left(2\right)}}$. That is, $x=0$. In addition, Theorem 25(ii) implies that the equations of the two directrices of the hyperbola are respectively given by

$$\begin{array}{c}\hfill y=\sqrt{\frac{-1-2+3}{2-(-1)+3}}x+\left[\frac{\sqrt{2-(-1)+3}\left(2\left(2\right)(-6)-\left(0\right)\left(0\right)\right)-\sqrt{-1-2+3}\left(2(-1)\left(0\right)-\left(0\right)(-6)\right)}{\sqrt{2-(-1)+3}\left(8\right)}\pm \frac{2}{(-1)\left(2+(-1)\right)+\left|3\right|}\sqrt{\frac{2(-12)}{-1-2-3}}\right],\end{array}$$

which is reduced to $y=-1$ and $y=-5$. Meanwhile, Theorem 27(ii) implies that the equations of the latera recta of the hyperbola are given by

$$\begin{array}{c}\hfill y=\sqrt{\frac{-1-2+3}{2-(-1)+3}}x+\frac{1}{8}\left[\frac{\sqrt{2-(-1)+3}\left(2\left(2\right)(-6)-\left(0\right)\left(0\right)\right)-\sqrt{-1-2+3}\left(2(-1)\left(0\right)-\left(0\right)(-6)\right)}{\sqrt{2-(-1)+3}}\pm 4\left(3\right)\sqrt{\frac{2(-12)}{-1-2-3}}\right],\end{array}$$

which is reduced to $y=0$ and $y=-6$. Note that in this case Theorems 26 and 28 cannot be applied.

Likewise, Corollary 5 implies that the general equations of the two degenerate cases of parabolas that respectively correspond to the directrices and the latera recta of the hyperbola in question are $y^2+6y+5=0$ and $y^2+6y=0$. Moreover, Corollary 20 implies that the equations of the two asymptotes of this hyperbola are given by

$$y=\left[\pm \frac{\sqrt{-2(-1)}}{(-1)}\right]x-\frac{\left(2\right)(-6)\mp \left(0\right)\sqrt{-2(-1)}}{2\left(2\right)(-1)},$$

which is reduced to $y=-\sqrt{2}x-3$ and $y=\sqrt{2}x-3$. Finally, it is clear that $\theta =0°$ in this case due to $B=0$. However, this fact can also be verified by Corollary 3, since $tan\theta =\sqrt{{\displaystyle \frac{-1-2+3}{2-(-1)+3}}}=0$, which guarantees $\theta =0°$. See Figure 12.

7. Concluding Remarks

7.1. Discussion on Rotated Conics

It is well-known that if $B=0$ in (3) (non-rotated conics), then the coefficients A and C determine what kind of conic section is represented by (3) and whether its focal axis is vertical or horizontal with respect to the non-rotated Cartesian axes. However, if $B\ne 0$ (rotated conics), then the nature of the conic section represented by (3) can be determined by Corollary 2 or Lemma 3 in general; moreover, they also work for the non-rotated conics.

On the other hand, if (3) corresponds to a rotated parabola or a rotated ellipse, then (10) and Corollaries 8, 9, 12, and 13 reveal that the relation between the sign of B and the signs of A and C, which are the same in these cases according to Corollary 2 and Lemmas 1(iii) and 2(i), determines whether the focal axis of the conic section in question is vertical or horizontal with respect to the Cartesian axes rotated by angle $\theta$. But if (3) corresponds to a rotated hyperbola, then (10) and Corollaries 18 and 19 reveal that the relation between the signs of B and $\Delta$ is what determines whether the focal axis of the rotated hyperbola is vertical or horizontal with respect to the rotated Cartesian axes, since $\mathrm{sgn}\left(A\right)$ is not necessarily equal to $\mathrm{sgn}\left(C\right)$ in this case.

Another implication of Lemmas 1(iii), 2(i), and 3(iii) and Corollary 2(iii) is that if $B\ne 0\ne \Delta$ whereas $A=0$ or $C=0$ in (3), then this equation corresponds to a rotated hyperbola. In addition, if $A=C=0$, then Remark 11 guarantees that this hyperbola is equilateral and is also rotated by $\theta =45°$, according to Theorem 1(i). Therefore, $A\ne 0\ne C$ with $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)$ and $B\ne 0\ne \Delta$ whenever (3) corresponds to rotated parabolas or rotated ellipses.

Likewise, although the formulae of (8) give the coordinates of the translated origin for any conic section with $d\ne 0$, these formulae are undefined for this purpose in the case of parabolas because $d=0$ in this case. So, Theorem 10 is what efficiently gives these coordinates for the case of parabolas, although this theorem requires the use of purely imaginary quantities whenever $\mathrm{sgn}\left(A\right)=\mathrm{sgn}\left(C\right)=-1$. However, Lemmas 1(iii), 2(i), and 3(ii) and Corollary 2(ii) guarantee that the imaginary number i will always be canceled during the process in this case, always obtaining real results in the end (see Example 2).

Finally, knowledge of the equations of the latera recta of any rotated conic section can facilitate the search for the coordinates of the extreme points of the corresponding latera recta, which can be obtained by solving non-linear systems of simultaneous equations involving the equation of the conic section in question and each linear equation of its latera recta, so this process implies solving quadratic equations only.

7.2. Conclusions

Nowadays, Analytic Geometry tends to be considered a finished topic in the big world of mathematics. However, the definition of conic radical (10) and Theorem 3 with all its implications for the study of conic sections have allowed us to develop a new vision of how everything about conic sections can be directly known from the six coefficients of (3), instead of using transformations based on trigonometric functions applied to the rotation angle on the Cartesian plane, which can be considered irrelevant from now on for a thorough analysis of conic sections. Therefore, Theorem 3 becomes a fundamental theorem for the study of conic sections.

The main contribution of this paper is that all the tools used in applied sciences to solve problems that require dealing with conic sections can be simplified a lot as a consequence of all the results presented here, which work by only applying basic arithmetic operations on the six coefficients of (3). In addition, our ultimate goal is to see our developments changing the perspective on how conic sections and Analytic Geometry in general have always been taught in high school and college courses at a global level.

So, we hope our findings serve the purpose of becoming an alternative to the traditional methodologies applied until now for the study of conic curves. Finally, all the results presented in this paper can be easily programmed to develop more effective software focused on applied sciences such as astronomy, physics and engineering, or even arts such as visual arts or architecture that require dealing with conic sections, due to the irrelevance of the rotation angle for the analysis of conic sections, which is introduced here.

The general formulae developed in this work constitute a substantial improvement over traditional analytic techniques for the study of rotated and non-rotated conic sections. Classical approaches rely heavily on computing the rotation angle

$$\theta =\frac{1}{2}arctan\left({\displaystyle \frac{B}{A-C}}\right),$$

followed by explicit trigonometric transformations, successive coordinate rotations, and case-by-case translations to eliminate the mixed term ($Bxy$) and reveal the intrinsic structure of the conic. These procedures are algebraically intensive, require multiple intermediate steps, and introduce the usual ambiguities associated with inverse trigonometric functions and angle selection. Moreover, standard textbooks do not provide general closed-form expressions for key geometric elements—such as vertices, foci, directrices, latera recta, or asymptotes—of arbitrarily oriented conics; instead, each type of conic must be treated separately after the axis-alignment process.

The framework presented in this paper addresses all these limitations simultaneously. By introducing the conic radical (10) and Theorem 3, we derive closed-form expressions for all fundamental geometric quantities directly in terms of the coefficients ($A,B,C,D,E,F$). These formulae apply uniformly to parabolas, ellipses, hyperbolas, and their degenerate counterparts, regardless of orientation, without requiring the explicit computation of ($\theta$) or any trigonometric functions. As a consequence, the rotation angle becomes mathematically irrelevant for the analysis: all canonical data of the conics are obtained instantaneously from their defining quadratic equation.

In particular, the general expressions for the vertices, foci, focal axes, latera recta, directrices, and asymptotes of any non-degenerate conic provide tools that are absent from the classical theory. Traditional methods offer such elements only for non-rotated standard forms, whereas the formulae developed here treat rotated and non-rotated cases on exactly the same footing. Furthermore, the canonical equations derived in Theorem 4 remove rotations and translations simultaneously, avoiding the usual combination of axis rotation, completion of squares, and coordinate shifts required in the classical approach. The method also incorporates a unified mechanism for detecting the type of the locus—ellipse, parabola, hyperbola, or degenerate case—via the conic radical criterion and the determinant ($\Delta$), thereby integrating classification and geometric computation into a single algebraic framework.

Overall, the results of this paper transform the analysis of conic sections into an algorithmic procedure governed solely by explicit algebraic expressions. This represents a significant conceptual and computational advance over existing approaches, offering a unified, rotation-free, and fully general methodology for the study of conics in the plane.