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 $S$ ?
 A: Hints:

*

*Translate the ellipsoid and the points so that the ellipsoid center is at the origin. This will greatly simplify the algebra.

*Minimize the sum of squared distances. This will give you the same point $C$, but you get rid of the square roots.

*The expression for $S$ is still going to be complicated, but you can find it’s minimum by setting it’s partial derivatives equal to zero. If you can’t solve the resulting equations algebraically, you’ll have to use numerical methods.

A: 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.
