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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

EDIT:

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.

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1 Answer 1

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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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