The foci and rotation angle of an ellipse, $E_0$, as a function of the coefficients of an equation of $E_0$

TL;DR

This paper derives formulas for the foci and rotation angle of an ellipse based on its algebraic equation coefficients, providing explicit relationships useful for geometric analysis.

Contribution

It introduces new explicit formulas linking ellipse foci and rotation angle directly to the coefficients of its defining equation.

Findings

Formulas for ellipse foci as functions of equation coefficients

Precise formula for ellipse rotation angle from coefficients

Theoretical foundation for geometric properties of ellipses

Abstract

First, we give a formula for the foci of an ellipse, $E_0$, as a function of the coefficients of an equation of $E_0$(see Theorem <ref>T2</ref>). To prove Theorem <ref>T2</ref>, we use two interesting formulas proven in <cite>B</cite> and in <cite>S</cite>. Our second result(see Theorem <ref>T3</ref>), is a more precise formula for the rotation angle of $E_0$ as a function of the coefficients of an equation of $E_0$.

Equations

$$x_c$$

$$y_c$$

$$x_c$$

$$y_c$$

$$A{x}^2+Bxy+C{y}^2+G=0\text{.}$$

$$\left\{\begin{array}[]{ll}x_{c}=c\cos\theta,y_{c}=c\sin\theta&\text{if }0\leq\theta\leq\dfrac{\pi}{2}\\ x_{c}=-c\cos\theta,y_{c}=-c\sin\theta&\text{if }\dfrac{\pi}{2}<\theta<\pi\end{array}\right.\text{.}$$

$$(a^2-x_c^2)x^2-2x_c{y}_{c}xy+(a^2-y_c^2)y^2-a^2{b}^2=0\text{.}$$

$$A=a^2-x_c^2,B=-2x_c{y}_c,C=a^2-y_c^2\text{,}$$

$$a^2$$

$$b^2$$

$$x_c$$

$$y_c$$

$$x_c$$

$$y_c$$

Full text

1 Introduction

The purpose of this note is two fold. First, we give a formula for the foci of an ellipse, $E_{0}$, as a function of the coefficients of an equation of $E_{0}$(see Theorem 2); To prove Theorem 2, we use two interesting formulas proven in [1] and in [3]. The main result in [1] expresses the foci of an ellipse, $E_{0}$, as a function of the coefficients of an equation of $E_{0}$, but also requires knowing the length of the major axis of $E_{0}$; We expand on that formula a little and give the proof here(see Theorem 1). A formula in [3] yields the length of the major axis of E_{0}\as a function of the coefficients(see Lemma 1). Theorem 1 and Lemma 1 then yield Theorem 2.

There are various ways to define the rotation angle, $\theta$, of a non–circular ellipse, $E_{0}$; Below we define $\theta$ to be the counterclockwise angle of rotation to the major axis of $E_{0}$ from the line thru the center of $E_{0}$ and parallel to the $x$ axis, with $0\leq\theta<\pi$; No matter how one defines $\theta$, it is always true that $\cot(2\theta)=\dfrac{A-C}{B}$; But what about a formula for $\theta$ itself ? Our second result(see Theorem 3), is a more precise formula for the rotation angle, $\theta$, of $E_{0}$, as a function of the coefficients of an equation of $E_{0}$. The latter formula was submitted as a correction for the previous formula for the rotation angle given in [2]. The formula given here now appears in [2] in a slightly different form. The proof of Theorem 3 then follows easily from the proof of Theorem 1.

While the formulas given in this note are undoubedtly known, and there are other ways of proving them, we found it interesting to use and highlight the results in [1] and in [3].

2 Foci as a Function of the Coefficients

Throughout, for a given ellipse, $E_{0}$, which is not a circle, we let $\theta$ denote the counterclockwise angle of rotation to the major axis of $E_{0}$ from the line thru the center of $E_{0}$ and parallel to the $x$ axis, with $0\leq\theta<\pi$; We let $(x_{0},y_{0})=$ center of $E_{0},a=$ length of semi–major and $b=$ length of semi–minor axes of $E_{0}$, respectively. Finally, we let $F_{2}=\left(x_{c},y_{c}\right)$ denote the rightmost focus of $E_{0}$(if $\theta=\dfrac{\pi}{2}$, we let $F_{2}$ denote the uppermost focus). Knowing $F_{2}$ easily yields the other focus, $F_{1}=\left(2x_{0}-x_{c},2y_{0}-y_{c}\right)$;

We now state an extension, and give a detailed proof, of the result in [1]. (i) gives the equation of an ellipse, $E_{0}$, given the foci of $E_{0}$, while (ii) gives the foci of $E_{0}$ given the equation of $E_{0}$. In each case, one must also know the length of the semi–major axis of $E_{0}$.

Theorem 1

Let $E_{0}$ be an ellipse which is not a circle, let $F_{2}=\left(x_{c},y_{c}\right)$ be the rightmost focus of $E_{0}$, and let $(x_{0},y_{0})$ be the center of $E_{0}$.

(i) Then the equation of $E_{0}$ can be written in the form $A\left(x-x_{0}\right)^{2}+B\left(x-x_{0}\right)\left(y-y_{0}\right)+C\left(y-y_{0}\right)^{2}-a^{2}b^{2}=0$, where $A=a^{2}-(x_{c}-x_{0})^{2},B=-2(x_{c}-x_{0})(y_{c}-y_{0})$, and $C=a^{2}-(y_{c}-y_{0})^{2}$.

(ii) If the equation of $E_{0}$ is written in the form $A\left(x-x_{0}\right)^{2}+B\left(x-x_{0}\right)\left(y-y_{0}\right)+C\left(y-y_{0}\right)^{2}-a^{2}b^{2}=0$, where $A,C>0$, then

[TABLE]

In addition, if $0\leq\theta<\dfrac{\pi}{2}$, then $B<0$, while if $\dfrac{\pi}{2}<\theta<\pi$, then $B>0$. Finally,

[TABLE]

Proof. It is clear that we may assume that $x_{0}=y_{0}=0$, so that the equation of $E_{0}$ has the form

[TABLE]

The implicit assumption in [1] is that $0\leq\theta<\dfrac{\pi}{2}$; We outline the proof in the case when $\dfrac{\pi}{2}\leq\theta<\pi$ as well. If $0\leq\theta<\dfrac{\pi}{2}$, then $F_{2}$ lies in quadrant 1, while if $\dfrac{\pi}{2}\leq\theta<\pi$, then $F_{2}$ lies in quadrant 4; Letting $c=\sqrt{a^{2}-b^{2}}$, we then have

[TABLE]

Recall that if $\theta=\dfrac{\pi}{2}$, then $F_{2}$ is the uppermost focus. Proceeding as in [1](we include the details here for completeness), we have:

$\sqrt{(x-x_{c})^{2}+(y-y_{c})^{2}}+\sqrt{(x+x_{c})^{2}+(y+y_{c})^{2}}=2a$, which implies that

$(x+x_{c})^{2}+(y+y_{c})^{2}=4a^{2}-4a\sqrt{(x-x_{c})^{2}+(y-y_{c})^{2}}+(x-x_{c})^{2}+(y-y_{c})^{2}$, and so

$a^{2}{\large(}(x-x_{c})^{2}+(y-y_{c})^{2}{\large)}=a^{4}-2a^{2}(xx_{c}+yy_{c})+(xx_{c}+yy_{c})^{2}$; Some simplification yields

$(a^{2}-x_{c}^{2})x^{2}-\allowbreak 2x_{c}y_{c}xy+(a^{2}-y_{c}^{2})y^{2}+a^{2}(c^{2}-a^{2})=0$, and using $b^{2}=a^{2}-c^{2}$ gives

[TABLE]

Matching (3) with (5) yields

[TABLE]

and $G=-a^{2}b^{2}$, which proves (i). To prove (ii): Note that $\left|x_{c}\right|=\sqrt{a^{2}-A}$ and $\left|y_{c}\right|=\sqrt{a^{2}-C}$.

Case 1: $B\neq 0$

Then $x_{c}\neq 0\neq y_{c}$ by (6), which implies that $0\neq\theta\neq\dfrac{\pi}{2}$ by (4); If $0<\theta<\dfrac{\pi}{2}$, then $x_{c}>0$ and $y_{c}>0$ by (4), which implies that $B<0,x_{c}=\sqrt{a^{2}-A}$, and $y_{c}=\sqrt{a^{2}-C}$; If $\dfrac{\pi}{2}<\theta<\pi$, then $x_{c}>0$ and $y_{c}<0$ by (4), which implies that $B>0,x_{c}=\sqrt{a^{2}-A}$, and $y_{c}=-\sqrt{a^{2}-C}$; That proves (1).

Case 2: $B=0$

Then $x_{c}=0$ or $y_{c}=0$ by (6); If $A<C$, then by (6) again, $a^{2}-x_{c}^{2}<a^{2}-y_{c}^{2}$, which implies that $y_{c}^{2}<x_{c}^{2}$, and so $y_{c}=0$; Thus $\theta=0$, which implies that $x_{c}=c>0$ by (4) and so $x_{c}=\sqrt{a^{2}-A}$; If $A>C$, then $a^{2}-x_{c}^{2}>a^{2}-y_{c}^{2}$, which implies that $y_{c}^{2}>x_{c}^{2}$, and so $x_{c}=0$; Thus $\theta=\dfrac{\pi}{2}$, which implies that$y_{c}=c>0$ by (4) and so $y_{c}=\sqrt{a^{2}-C}$; That proves (2).

For the following two lemmas, we let $\Delta=4AC-B^{2}$ and $\delta=CD^{2}+AE^{2}-BDE-F\Delta$. The following result can be found in [3].

Lemma 1

Suppose that $E_{0}$ is an ellipse with equation $Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$; Let $a$ and $b$ denote the lengths of the semi–major and semi–minor axes, respectively, of $E_{0}$, and let $\mu=\dfrac{4\delta}{\Delta^{2}}$. Then

[TABLE]

We state the following useful general lemma about equations of ellipses. The second condition ensures that the conic is non–degenerate, while the first condition ensures that the conic is an ellipse.

Lemma 2

The equation $Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$, with $A,C>0$, is the equation of an ellipse if and only if $\Delta>0$ and $\delta>0$.

Using Lemma 1, we are now able to give a formula for the foci of $E_{0}$, given an equation of $E_{0}$, without knowing the length of the semi–major axis of $E_{0}$, as with Theorem 1.

Theorem 2

Let $E_{0}$ be an ellipse which is not a circle, and let $r=\sqrt{(A-C)^{2}+B^{2}}$; Let $F_{2}=\left(x_{c},y_{c}\right)$ be the rightmost focus of $E_{0}$ and let $(x_{0},y_{0})$ be the center of $E_{0}$. If the equation of $E_{0}$ is written in the form $A\left(x-x_{0}\right)^{2}+B\left(x-x_{0}\right)\left(y-y_{0}\right)+C\left(y-y_{0}\right)^{2}-a^{2}b^{2}=0$, where $A,C>0$, then

[TABLE]

[TABLE]

Remark 1

To use Theorem 2, one must first rewrite the equation of $E_{0}$ so that it has the form given in Theorem 2. First one writes the equation of $E_{0}$ in the form $A\left(x-x_{0}\right)^{2}+B\left(x-x_{0}\right)\left(y-y_{0}\right)+C\left(y-y_{0}\right)^{2}+F=0$ using the formula $x_{0}=\dfrac{BE-2CD}{\Delta}$, $y_{0}=\dfrac{BD-2AE}{\Delta}$; One can then obtain $a^{2}b^{2}$ without needing to know $a^{2}$ or $b^{2}$ since it follows easily by Lemma 1 that $a^{2}b^{2}=\dfrac{\Delta}{\delta}=\dfrac{4AC-B^{2}}{CD^{2}+AE^{2}-BDE-F(4AC-B^{2})}$, where $Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$ is any given equation of $E_{0}$. Multiplying thru by $\dfrac{-a^{2}b^{2}}{F}$ then yields the proper form.

Proof. As in the proof of Theorem 1, we may assume, without loss of generality, that $x_{0}=y_{0}=0$, so that the equation of $E_{0}$ has the form $Ax^{2}+Bxy+Cy^{2}-a^{2}b^{2}=0$. By Lemma 2, $\Delta\neq 0$, and by Lemma 1, with $D=E=0$ and $F=-a^{2}b^{2}$, we have $\delta=a^{2}b^{2}\Delta$, which implies that $\mu=\dfrac{4a^{2}b^{2}\Delta}{\Delta^{2}}=\dfrac{4a^{2}b^{2}}{\Delta}$; Also by Lemma 1, $a^{2}b^{2}=\dfrac{\mu^{2}}{4}{\large(}(A+C)^{2}-(A-C)^{2}-B^{2}{\large)}=\left(\dfrac{4\delta^{2}}{\Delta^{4}}\right)\Delta=\dfrac{4\delta^{2}}{\Delta^{3}}$; Thus $\dfrac{\delta}{\Delta}=a^{2}b^{2}=\dfrac{4\delta^{2}}{\Delta^{3}}$, and so $\delta=\dfrac{1}{4}\Delta^{2}$; Hence $\mu=\dfrac{4\delta}{\Delta^{2}}=1$, which implies, by Lemma 1, that $a^{2}=\dfrac{A+C+r}{2}$. Substituting $a^{2}-A=\dfrac{r+C-A}{2}$ and $a^{2}-C=\dfrac{r+A-C}{2}$ into Theorem 1 yields Theorem 2.

3 Rotation Angle

Below we give a formula for the rotation angle, $\theta$, of a non–circular ellipse, $E_{0}$, as a function of the coefficients of an equation of $E_{0}$. We also give a simple formula for $\tan\theta$. Here we are assuming that $0\leq\cot^{-1}x<\pi$.

Theorem 3

Let $E_{0}$ be an ellipse with equation $Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$, with $A,C>0$; Let $\theta$ denote the counterclockwise angle of rotation to the major axis of $E_{0}$ from the line thru the center of $E_{0}$ and parallel to the $x$ axis, with $0\leq\theta<\pi$;Let $r=\sqrt{(A-C)^{2}+B^{2}}$.

(i) \theta=\left\{\begin{array}[]{ll}(1+\mathop{\mathrm{s}gn}B)\dfrac{\pi}{4}+\dfrac{1}{2}\cot^{-1}\left(\dfrac{A-C}{B}\right)&\text{if }B\neq 0\\ {\large(}1+\mathop{\mathrm{s}gn}(A-C){\large)}\dfrac{\pi}{4}&\text{if }B=0\end{array}\right. and

(ii) $\tan\theta=\dfrac{B}{A-C-r}$ if $B\neq 0$ or $B=0$ and $A<C$.

Remark 2

Note that if $A=C$ and $B=0$, we have a circle and hence no rotation angle.

Proof. Again, we assume that $x_{0}=y_{0}=0$, where $E_{0}$ has center $=(x_{0},y_{0})$; By (6) and (4), $A-C=y_{c}^{2}-x_{c}^{2}=c^{2}\sin^{2}\theta-c^{2}\cos^{2}\theta=(a^{2}-b^{2})(\sin^{2}\theta-\cos^{2}\theta)=(b^{2}-a^{2})\cos(2\theta)$, and $B=-2x_{c}y_{c}=-2c^{2}\cos\theta\sin\theta=(b^{2}-a^{2})\sin(2\theta)$ for any $0\leq\theta\leq\pi$; We use the well–known formula $\cot(2\theta)=\dfrac{A-C}{B}$, which implies that $\theta=\dfrac{1}{2}\cot^{-1}\left(\dfrac{A-C}{B}\right)$ or $\theta=\dfrac{\pi}{2}+\dfrac{1}{2}\cot^{-1}\left(\dfrac{A-C}{B}\right)$, depending upon whether $\theta$ lies in quadrant 1 or quadrant 2. Note that $b^{2}-a^{2}<0$;

Case 1: $B\neq 0$

If $B<0$, then $(b^{2}-a^{2})\sin(2\theta)<0$, which implies that $\sin(2\theta)>0$ and so $0<\theta<\dfrac{\pi}{2}$; Thus $\theta=\dfrac{1}{2}\cot^{-1}\left(\dfrac{A-C}{B}\right)$; If $B>0$, then $(b^{2}-a^{2})\sin(2\theta)>0$, which implies that $\sin(2\theta)<0$ and so $\dfrac{\pi}{2}<\theta<\pi$; Thus $\theta=\dfrac{\pi}{2}+\dfrac{1}{2}\cot^{-1}\left(\dfrac{A-C}{B}\right)$.

**Case 2: $B=0$ **

Then $\sin(2\theta)=0$, which implies that $\theta=0$ or $\theta=\dfrac{\pi}{2}$; If $A>C$, then $A-C=(b^{2}-a^{2})\cos(2\theta)>0$, which implies that $\cos(2\theta)<0$ and so $\theta=\dfrac{\pi}{2}$; If $A<C$, then $A-C=(b^{2}-a^{2})\cos(2\theta)<0$, which implies that $\cos(2\theta)>0$ and so $\theta=0$. That proves (i).

While one could use the fact that $\cot(2\theta)=\dfrac{A-C}{B}$, we find it easier to proceed as follows to prove (ii). Now by (4), $\tan\theta=\dfrac{y_{c}}{x_{c}}$

Case 1: $B\neq 0$

Then by Theorem 2, $x_{c}=\sqrt{(r+C-A)/2}$ and $y_{c}=-(\mathop{\mathrm{s}gn}B)\sqrt{(r+A-C)/2}{\large)}$, which implies that

$\tan\theta=\dfrac{-(\mathop{\mathrm{s}gn}B)\sqrt{(r+A-C)/2}{\large)}}{\sqrt{(r+C-A)/2}}=-(\mathop{\mathrm{s}gn}B)\sqrt{\dfrac{r+A-C}{r+C-A}}$; Now $\dfrac{r+A-C}{r+C-A}$ simplifies to $\dfrac{B^{2}}{{\large(}r-(A-C){\large)}^{2}}$, and $r-(A-C)=\sqrt{(A-C)^{2}+B^{2}}-(A-C)>0$ using the inequality $\sqrt{x^{2}+y^{2}}-x>0$ for any $x$; Thus $\tan\theta=-(\mathop{\mathrm{s}gn}B)\sqrt{\dfrac{B^{2}}{{\large(}r-(A-C){\large)}^{2}}}=\dfrac{-(\mathop{\mathrm{s}gn}B)\left|B\right|}{r-(A-C)}=\dfrac{-B}{r-(A-C)}=\dfrac{B}{A-C-r}$;

Case 2: $B=0$

If $A<C$, then by Theorem 2, $x_{c}=\sqrt{(r+C-A)/2}$ and $y_{c}=\sqrt{(r+A-C)/2}{\large)}$, and the rest follows as above.

4 Example

Consider the ellipse with equation $\allowbreak 4x^{2}+2xy+6y^{2}-6x+10y=1$; Using $A=4$, $B=2$, $C=6$, $D=-6$, and $E=10$, one has $\Delta=4(4)(6)-2^{2}=\allowbreak 92$ and $\delta=\allowbreak 828$; Using Remark 1 yields $a^{2}b^{2}=\dfrac{4\delta^{2}}{\Delta^{3}}=\allowbreak\dfrac{81}{23}$ and $x_{0}=\allowbreak 1$, $y_{0}=\allowbreak-1$; Rewriting the equation gives $4(x-1)^{2}+2(x-1)(y+1)+6(y+1)^{2}-9=0$; Multiplying thru by $\dfrac{-a^{2}b^{2}}{-9}=\dfrac{9}{23}$ yields $\dfrac{36}{23}(x-1)^{2}+\dfrac{18}{23}(x-1)(y+1)+\dfrac{54}{23}(y+1)^{2}-\dfrac{81}{23}=0$; Now we have $A=\dfrac{36}{23}$, $B=\dfrac{18}{23}$, $C=\dfrac{54}{23}$, and $r=\allowbreak\dfrac{18\sqrt{2}}{23}$; By Theorem 2, $F_{2}=\allowbreak\left(1+\dfrac{3}{23}\sqrt{23+23\sqrt{2}},-1-\dfrac{3}{23}\sqrt{-23+23\sqrt{2}}\right)$, which implies that $F_{1}=\left(1-\dfrac{3}{23}\sqrt{23+23\sqrt{2}},-1+\dfrac{3}{23}\sqrt{-23+23\sqrt{2}}\right)\allowbreak$; By Theorem 3, $\theta=(1+\mathop{\mathrm{s}gn}B)\dfrac{\pi}{4}+\dfrac{1}{2}\cot^{-1}\left(\dfrac{A-C}{B}\right)=\allowbreak\dfrac{7\pi}{8}$; One can verify that $\tan\theta=\dfrac{B}{A-C-r}=\allowbreak 1-\sqrt{2}\allowbreak$;