It is known that if there is an elliptical billiard table and we shoot a ball from one focus of the ellipse in any direction, it is necessary that the ball reaches the other focus in only one bounce (or no bounce).
You can see here that the angles $\beta$ are the same. I wanted to prove this, so using implicit differentiation I calculated the slope of the line F1P (called $l$), the slope of the line F2P (called $m$), and the slope of the line nP (called $n$). Then I got the following formula, for $F_1(-\alpha,0)$ and $F_2(\alpha, 0)$ and $P(x,y)$:
$$l = \frac{y}{x+\alpha}$$ $$m = \frac{a^2}{b^2}\frac{y}{x}$$ $$n = \frac{y}{x-\alpha}$$
I wanted to verify that the both $\beta$ are the same by comparing the tangent values:
$$\frac{m-l}{1+ml} = \tan(\beta) = \frac{n-m}{1+nm}$$
Here I used the formula $$\tan (\theta -\phi)=\frac{\tan(\theta)-\tan(\phi)}{1+\tan(\theta)\tan(\phi)}$$
I put everything into the Computer Algebra System and I get the following result:
$$\frac{\alpha^2(x+a^2)y}{a^2b^2+\alpha^2b^2x} = \frac{-\alpha^2(x-a^2)y}{a^2b^2-\alpha^2b^2x}$$
which is not an identity. This contradicts with the obvious result that we can guarantee (that $\beta$ are the same). What is wrong?
We take an ellipse centered at origin with major semi-axis $a$ and minor semi-axis $b$.
$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$
Focii of ellipse are $F_1 (- \alpha, 0)$ and $F_2 (\alpha, 0)$, where $\alpha^2 = a^2 - b^2$
We take a point $P$ on ellipse as $(a \cos\theta, b \sin\theta)$
Then slopes of lines $PF_1$ and $PF_2$ are,
$m_1 = \dfrac{b \sin \theta - 0}{a \cos\theta + \alpha}, m_2 = \dfrac{b \sin \theta - 0}{a \cos\theta - \alpha}$
Slope of normal line passing through point $P$ is,
$m = \dfrac{a \ sin\theta}{b \cos\theta}$
Now there are two ways to proceed - show that one of the angle bisectors of the lines $PF_1$ and $PF_2$ has the slope $m$. The other way is to show that the angle between normal line and $PF_1$, and the angle between normal line and $PF_2$ are same. I will use the second approach as I think the working is a bit simpler in this approach.
$m - m_1 = \dfrac{a \ sin\theta}{b \cos\theta} - \dfrac{b \sin \theta}{a \cos\theta + \alpha} = \dfrac{(a^2-b^2) \sin \theta \cos\theta + a \alpha \sin \theta}{b \cos\theta (a \cos\theta + \alpha)} = \dfrac{\alpha \sin \theta (a + \alpha \cos\theta)}{b \cos\theta (a \cos\theta + \alpha)}$
(using $a^2 - b^2 = \alpha^2$)
$1 + m \ m_1 = \dfrac{b ( a + \alpha \cos\theta)}{b \cos\theta (a \cos\theta + \alpha)}$
So angle $\theta_1$ between line going through $P$ and focus $(-\alpha, 0)$, and the normal line at $P$ is given by,
$\tan \theta_1 = \left| \dfrac{m-m_1}{1 + m \ m_1} \right| = \left|\dfrac{\alpha \sin \theta}{b}\right|$
Similarly show that $\tan \theta_2 = \left|\dfrac{\alpha \sin \theta}{b}\right|$
Can you take it from here?
Alternate approach: make use of angle bisector theorem (wiki). If normal line intersects $F_1F_2$ at point $Q$, show that
$\dfrac{PF_1}{PF_2} = \dfrac{QF_1}{QF_2}$ ...$(i)$
If point $P$ is $(r, s)$, equation of normal line is
$ \displaystyle y - s = \frac{a^2s}{b^2r} (x-r)$ so point $Q$ is $\left (\dfrac{\alpha^2 r}{a^2}, 0\right)$
Now just take distances between points and you can easily show $(i)$.