Reflection across an Ellipsoid A reflective ellipsoid is given by
$$ \dfrac{x^2}{16} + \dfrac{y^2}{9} + \dfrac{(z - 2)^2}{4} = 1 $$
A light source, emitting rays in all directions, is placed at $A=(10,4,3)$.  Find the point $C=(x,y,z)$ on the surface of the ellipsoid that minimizes the sum of distances $AC + CB$, where $B = (8, 12, 7) $
My attempt:
Parameterize the equation of the ellipsoid as follows
$ x = 4 \sin \theta \cos \phi $
$ y = 3 \sin \theta \sin \phi $
$ z = 2 + 2 \cos \theta $
Now the sum of the two distances is
$S = AC + CB = \sqrt{ (4 \sin \theta \cos \phi - 10)^2 + (3 \sin \theta \sin \phi - 4)^2 + (2 + 2 \cos \theta - 3)^2 } \\
+\sqrt{ (4 \sin \theta \cos \phi - 8)^2 + (3 \sin \theta \sin \phi - 12)^2 + (2 + 2 \cos \theta - 7)^2 }$
How can I minimize $S$ ?

Take a generic point on the ellipsoid, $(a,b,c)$ (first condition on $a,b,c$). To mimimize your sum, you need the ray of light going from $A$ to $B$ bouncing at the ellipsoid to be reflected at $C$. The normal to the ellipsoid at $(a,b,c)$ is the gradient $(a/8, 2b/9, (c-2)/2)$. You need two additional conditions: the angles formed between $CA$ and $CB$ and the normal vector should be equal and the three vectors should be coplanar (their triple product equal to zero). Solve the system of three equations.
