I am trying to derive a relation between the angle from the center of ellipse $C$ and its true anomaly (angle from the focal point $F$) (alpha vs. beta in the picture), for a general ellipse. For some reason, it seems I am doing some mistake somewhere.

What I tried is this. When we take the polar description with regard to the center

$y = r\sin(\alpha) = \frac{ab}{\sqrt{(b\cos\alpha)^2+(a\sin\alpha)^2}}\sin\alpha$

Also for the other equation from the focus point we have $y=r\sin(\beta)$, were $r$ is the length with respect to the focus point, i.e.,

$r=\frac{a(1-e^2)}{1\pm e\cos\beta}.$

By putting the equations together and squaring, I get a monster equation:

$\frac{a^2\sin^2\alpha}{(b\cos\alpha)^2+(a\sin\alpha)^2} = \frac{b^2\sin^2\beta}{1\pm2e\cos\beta+e^2\cos^2\beta}$

This leads to a horrible quadratic equation in $\sin\alpha$. Am I doing some error somewhere. Is there some other option, some "shortcut" to derive it in an easier way?

Any hints are welcome.

enter image description here


After fighting with the equations, I got this result:

$\sin\alpha = \pm\frac{b}{a}q\sin\beta\sqrt{\frac{1}{1-e^2q^2\sin^2\beta}}, \to \alpha = \arcsin\left(\pm\frac{b}{a}q\sin\beta\sqrt{\frac{1}{1-e^2q^2\sin^2\beta}}\right)$

where $q=\frac{b}{a}\frac{1}{1\pm e\cos\beta}$ and $e=\sqrt{1-\left(\frac{b}{a}\right)^2}$.

Doing numerical testing, it looks like it is a correct result, except for angles larger than $90^°$, where some sign error seems to be. If there is a simpler and nicer solution, which works for any angle $\beta$, I would be grateful to see it.


1 Answer 1


This answer deals with the case where $\beta\in[0,\pi]$.

If $\beta\not=\frac{\pi}{2}$, then the intersection point $(X,Y)$ where $Y\ge 0$ of $y=\tan\beta\ (x-ae)$ with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is given by $$(X,Y)=\left(\frac{-b^2c^2ae +abc\sqrt{D}}{A}+ae,\frac{-b^2scae +abs\sqrt{D}}{A}\right)$$

where $s:=\sin\beta,c:=\cos\beta,A:=a^2s^2+b^2c^2,D:=a^2s^2+b^2c^2-a^2s^2e^2$.

Therefore, we get $$\alpha=\begin{cases} \arctan\left(\frac YX\right)=\arctan\left(\frac{-b^2sce+bs\sqrt D}{a^2s^2e+bc\sqrt D}\right) & \text{if $\beta\in\big[0,\frac{\pi}{2}\big)\cup\big(\frac{\pi}{2},\pi-\arctan\left(\frac{b}{ae}\right)\big)$} \\\\ \pi-\arctan\left(\frac{Y}{-X}\right)=\pi-\arctan\left(\frac{b^2sce-bs\sqrt D}{a^2s^2e+bc\sqrt D}\right) & \text{if $\beta\in\big(\pi-\arctan\left(\frac{b}{ae}\right),\pi\big]$}\\\\ \arctan\left(\frac{b\sqrt{1-e^2}}{ae}\right)& \text{if $\beta=\frac{\pi}{2}$}\\\\ \frac{\pi}{2}&\text{if $\beta=\pi-\arctan\left(\frac{b}{ae}\right)$} \end{cases} $$


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