Finding the image in an elliptic mirror I'm trying to find the location of the image of a point being reflected by a mirror shaped like (half of an) ellipse. The goal is to find a transformation $\mathbb{R}^2\rightarrow \mathbb{R}^2$ which sends each point on one side of the ellipse to the location of its mirror image point on the other side. (Points on the boundary of the ellipse remain fixed.)
Previously, I've successfully found this transformation for parabolic mirror. The transformation is 
$$\widehat{x} = \frac{1+\sin{\alpha}}{\cos{\alpha}}\\ \widehat{\alpha} = \frac{1/4x^2-f}{x}$$
which you can prove using the following diagram. The coordinate system is chosen to take advantage of the parabola's reflective property: it sends vertical lines to lines through the focus and vice-versa; this is evident in the way that $\widehat{x}$ only depends on $\alpha$ and $\widehat{\alpha}$ only depends on $x$. 
I thought it would be similarly straightforward to compute this transformation for an ellipse. I expected that here a convenient choice of coordinate system would be the angle formed between the target point and each focus—but when I attempt to use the ellipse's reflective property (rays from one focus reflect into the other focus) and solve for the location of the image point, I get increasingly hairy math involving quadratic equations (where the ray from one focus toward the point intersects the boundary of the ellipse) and nothing seems to simplify. 
I expected elegance similar to the parabola—am I taking a wrong approach somehow? (And if not, what's the explanation for the difference?)

Edit: I did find another almost-convenient coordinate system $\langle \theta_1, \theta_2\rangle$; for a given point, to compute $\theta$, draw a ray from the point to one focus. Find the point where the ray intersects the ellipse. The angle between that intersection point and the origin is $\theta$. Doing so for both foci gives two angles. By the reflection property of the ellipse, the reflection map simply exchanges the angles $\theta_1 \leftrightarrow \theta_2$.
This is mathematically simple, but unsatisfying to me for a few reasons. First, the angles $\theta$ seem unnatural because they involve computing a point of intersection, which feels like a complicated intermediate variable. (I wish we could solve in terms of more natural-seeming parameters such as the angles between the foci and the point— which might still be difficult, because there are many confocal ellipses and so some intersection with the boundary must be involved somewhere.) Second, it doesn't seem to obviously relate to the formula I got for the parabola, whereas I would expect the parabola formula to emerge as an extreme limiting case of the ellipse.


 A: Let the point to be reflected be $P(u, v)$.
Equn of ellipse is
$$ E: x^2/a^2 + y^2/b^2 = 1$$
where $ y>=0 $ since you mentioned half mirror.
To get image $ P'(u', v') $ of P we need to drop a normal from P onto E. Let the normal intersect E at $ Q(a \cos q, b \sin 
 q)$. Then we can write the equn of normal at Q as:
$$ a x \sec q - b y \csc q = a^2 - b^2 $$
Since it passes through P, we have
$$ a u \sec q - b v.\csc  q = a^2 - b^2 \tag1 $$
now obviously Q is mid point of P and P'. So that
$$ a \cos q = u + u', \quad 2a \sin q = v + v',$$
or
$$(u', v')= (2a \cos q - u, 2a \sin q - v)  \tag2 $$
Now, all you need to do is to find $(\cos q , \sin q)$ from (1) which itself is an equation in q and substitute those values in (2).
However, here comes the important bit: for a full ellipse, if you consider a point inside the ellipse, then there are
four normals that can be drawn from it to the ellipse in general.
Hence for a half ellipse there can be either 1, 2, 3 images of the same point. There wont be a unique image. Hence all those images can be found be different combinations of $(\cos q , \sin q ) $ we obtained from (1).
Further we can write
$$ (\cos q , \sin q) = [(1 - t^2)/(1+t^2) , 2t/(1+t^2) ]$$
respectively where $t= \tan(q/2)$
Plugging this into (1) shall render a quartic polynomial equn with max 4 real roots. Out of which you will need to eliminate the ones for which the y coordinate $ b \sin q<0 $ or $ \sin q<0 .$
