I'm trying to prove the reflection property of the ellipses for an optics problem. The property is that that a ray of light originated at one of the ellipse's foci reflects in such a way to pass through the other focus.
The simplest way I figured out to prove this is to prove that a line connecting one of the focus to any point of the ellipse has the same angle with the normal at that point as a line connecting the other focus with the same point.
So... Let
$\frac{\displaystyle x^2}{\displaystyle a^2}+\frac{\displaystyle y^2}{\displaystyle b^2}=1\,\,\,\rightarrow\,\,\,y=\pm b\sqrt{1-\frac{\displaystyle x^2}{\displaystyle b^2}}$
be the ellipse equation. The foci are at positions $(0,\pm f)$ where $f=\sqrt{b^2-a^2}$.
The lines connecting the foci to the point $(x_{0},y_{0})$ have equations:
$line\,1\,\,\,y=y_{0}+\left(\frac{\displaystyle y_{0}-f}{\displaystyle x_{0}}\right)(x-x_{0})$
$line\,2\,\,\,y=y_{0}+\left(\frac{\displaystyle y_{0}+f}{\displaystyle x_{0}}\right)(x-x_{0})$
and the line equation of the normal to the point $(x_{0},y_{0})$ is
$y=y_{0}-\frac{\displaystyle 1}{\displaystyle y'(x_{0})}(x-x_{0})$ $=y_{0}+\frac{\displaystyle a^2}{\displaystyle b^2}\frac{\displaystyle y(x_{0})}{\displaystyle x_{0}}(x-x_{0}).$
The angle between line 1 and the normal is then
$\theta_{1}=\arctan\left(\frac{\displaystyle a^2}{\displaystyle b^2}\frac{\displaystyle y_{0}}{\displaystyle x_{0}}\right)-\arctan\left(\frac{\displaystyle y_{0}-f}{\displaystyle x_{0}}\right)$
And the angle between line 2 and the normal is
$\theta_{2}=\arctan\left(\frac{\displaystyle y_{0}+f}{\displaystyle x_{0}}\right)-\arctan\left(\frac{\displaystyle a^2}{\displaystyle b^2}\frac{\displaystyle y_{0}}{\displaystyle x_{0}}\right)$
And so... I'm trying to proof (with no success still) that $\theta_{1}=\theta{2}$.
Could you lend me a tip?
Thank you very much.
A hint:
From the formula for $\tan(\alpha-\beta)$ one can deduce that $$\arctan v-\arctan u=\arctan{v-u\over 1+uv}\ ,$$ up to multiples of $\pi$. The latter should be taken care of, if necessary.