How to determine the reflection point on an ellipse Here is my problem. There are  two points P and Q outside an ellipse, where the coordinates of the P and Q are known. The shape of the ellipse is also known. A ray comming from point P is reflected by the ellipse and arrivates at Q. The question is how to determine the reflection point on the ellipse. I mean is there any analytical method to calculate the coordinate of the reflection point?


One possible solution can be as follows:
Let $\rm P{(x_1,y_1)} \ Q{(x_2,y_2)}, and\  R{(h,k)}, $ be the points.
Let ellipse ($\rm S$) be a special ellipse with centre (0,0)
$$\begin{align}\rm S &=\rm \dfrac{x^2}{a^2}+ \dfrac{y^2}{b^2} = 1 \\
\rm Equation \ of \ tangent\ at\ R &:\rm \dfrac{xh}{a^2}+ \dfrac{yk}{b^2}-1 = 0  \ \ \ \ \ \ ... (i)\end{align}$$
Since the line segment $\rm PR$ (slope $\rm m_2$) is equally inclined to the tangent ( slope $\rm m_1$) as the segment $\rm QR$ ( slope $\rm m_3$), we can equalise the angles.

$$\rm\tan\theta=\dfrac{m_1 - m_2}{1+m_1m_2}$$
$$\rm\tan\phi=\dfrac{m_1 - m_3}{1+m_1m_3}$$
$$\rm \implies \tan\theta = \tan\phi \ \ \ \ \ \ ... (ii)$$
Find $\rm m_1, m_2, m_3$
You'll get $2$ equations with $2$ unknowns $\rm h$ and $\rm k$.

Note In my answer, I used eqn of standard ellipse with centre at origin. But this method will work for any ellipse. For finding tangent eqn, just put $T=0$

EDIT
Here is a graph which shows tangent, and a line inclined equally to it (it tangent is the angle bisector of the lines.) : Desmos
