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?

 A: Hint:
We can use a method similar to that of Max Payne, but representing the ellipse by its parametric equations $x = a \cos t$, $y = b \sin t$. By differentiation, the tangent vector is $(- a \sin t, b \cos t)$, hence the normal vector $(b \cos t, a \sin t)$.
Now we express the equality of the angles to the normal, by the equality of dot products
$$\frac{(P_x - a \cos t) b \cos t + (P_y - b \sin t) a \sin t}{\sqrt{(P_x - a \cos t)^2+(P_y - b \sin t)^2}} = \frac{(Q_x - a \cos t) b \cos t + (Q_y - b \sin t) a \sin t}{\sqrt{(Q_x - a \cos t)^2+(Q_y - b \sin t)^2}}.$$
This can be rewritten as a nasty trigonometric polynomial of degree four by squaring and reducing to the common denominator, then rationalized to an octic polynomial.
A: You could orient the coordinate system to prefer the ellipse. That is, take your original coordinate system, and find the translation that moves the ellipse origin to the center. And find the rotation that aligns the ellipse axes with the coordinate axes. Apply these to $P$ and $Q$, so now the problem is...
We have an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and points $P'=(h,k)$ and $Q'=(m,n)$. Where is the point $M'$ on the ellipse such that the ray $P'M'$ reflects to $Q'$?
The ellipse edge is traced by $r(t)=(a\cos(t),b\sin(t))$. At $t$, the tangent vector is $\vec{u}=\langle -a\sin(t),b\cos(t)\rangle$.
At $t$, the line from $P'$ has direction $\vec{v}=\langle a\cos(t)-h,b\sin(t)-k\rangle$.
At $t$, the line from $Q'$ has direction $\vec{w}=\langle a\cos(t)-m,b\sin(t)-n\rangle$.
The relationship between these three vectors needs to be that the sum of normalized $\vec{v}$ and normalized $\vec{w}$ is orthogonal to $\vec{u}$. That is: $$\left(\frac{\vec{v}}{\lVert\vec{v}\rVert}+\frac{\vec{w}}{\lVert\vec{w}\rVert}\right)\cdot \vec{u}=0$$
That sets up a polynomial equation in $\sin(t)$ and $\cos(t)$, where you could solve for $t$. There may be multiple solutions, some of which correspond to a ray that passes into the ellipse and reflecting from the inside. After finding $t$, that tells you $M'$. And then you can undo the original coordinate transformation to answer the original question.
A: 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
A: To get a comprehensive picture we have to find coordinates of reflection point $ M (x,y)$
Elemetary vector approach adopted as follows. 
Vector $MP \rightarrow (x_p- x)  + i (y_p- y) ; $
Vector $M1  \rightarrow  (x-c,y )  + i y ; $
Vector $MQ\rightarrow (x_q- x)  + i (y_q- y) ; $
Vector $M1\rightarrow  (x+c ,y )  + i y ; $
Angle between vectors $(MQ,M1)$ = angle between vectors $(MP,M2)$; 
where $(1,2)$ are abbreviated for $(F1,F2)$
So equating their cosines through dot product we get locus of all required points of incidence sought by the OP for an entire family of confocal ellipses ( for that matter all conics, when appropriate sign changes are made.)
$$ \cos \theta_q = \dfrac{ (x_q- x) (x+c) - y (y_q- y) } { \sqrt{(x_q-x)^2 + (y_q-y)^2 } \cdot  \sqrt{(x+c)^2 + y^2} }$$
$$ \cos \theta_p = \dfrac{ (x_p- x) (x-c) - y (y_p- y) } { \sqrt{(x_p-x)^2 + (y_p-y)^2 } \cdot  \sqrt{(x-c)^2 + y^2} }$$
Equation of locus of confocal ellipses used to plot is
$$ \cos \theta_q = \mp \cos \theta_p $$  
However the reflection point is available only implicitly. Eighth degree (Octic?) polynomial.
Given points are $P,Q,F1,F2 $ .Wlog ellipse has x-axis horizontal. Constants used to graph ( in Mathematica ContourPlot ) below:
$$ [ (x_p,y_p),(x_q,y_q), (a,b,c)]= [ ( -4.8, 2.9), (3., 4.1), (5,3,4)] $$
Only one value of $c=4$ is represented on magenta colored ellipse:
Each square has side length = 5. 
Both convex and concave reflection points ( purple and red loci ) are obtained graphically by the method here. It is possible to numerically get the reflection points by  solving the locus and ellipse profiles together (not given here).
EDIT 1:
The scheme works also when given points are inside a bigger confocal ellipse (OP did not ask for this option). Here $ (a,c) = (8,4). $  Pictured around the same rope and hump loci, for same positions of $ P,Q, F_1, F_2 $. Colors interchanged. 

A: EDIT 1:
In other words, when the position of a point on ellipse M is correct we note that two conditions need to be satisfied:
Using  minimization of length by Fermat principle and constraint condition $( P M F_2)$ is a straight line, [$M (x,y)$ a point on ellipse, $F_2$ is focal point on the side opposite of normal in a reflection setting ] we can use a Lagrangian
$$  F(x,y)= \sqrt{(x_p- x )^2  + (y_p- y)^2} + \sqrt{(x_q- x)^2 + (y_q- y)^2} ; $$
$$ G(x,y) = \frac{y_p-y}{x_p-x} - \frac{y_ {F_2}-y}{x_{F_2}-x}, $$
since position of focus $F_2$ is known to be on a straight line i.e., in the solution we should have lines $ PMF_2, QMF_1 $ straight (and later on would be shown equally inclined to the ellipse tangent at $M$),
we should satisfy $$ \frac{F_x}{F_y} = \frac{G_x}{G_y} = -\lambda. $$
EDIT 2:
The following interesting property can be used for a possible total solution.
Just as $M$ is tangent point of contact of the given ellipse with foci $F_1,F_2$  that contains contact incident/reflected rays $PF_2,Q F_1, $ so also is  $M$ is a common tangent point of contact of a dual ellipse with foci $P,Q $  sharing the same incident/reflected rays $PF_2,Q F_1. $
This leads to an unexpected conclusion that straight line$ P F_2, Q F_1,$ and the tangent at M (red line), should be concurrent at the tangent point $M$ and more importantly also that the
positions of $P,Q$ cannot be be assigned arbitrarily as given/fixed. If P is given, point Q must necessarily lie on straight line  $F_1 M $ .
EDIT 3:
Schematic dual ellipses pictured in addition to the existing image.
A laser beam focused from P onto ellipse goes through Q, obeying reflection laws, which can be shown by Fermat principle. Likewise from F1 to F2 internally there is a reflection.As vertically opposite angles must be equal, $ PMQ $ must be a straight line.

