An ellipsoid with equation
$ (\mathbf{r} - \mathbf{r_0} )^T Q (\mathbf{r} - \mathbf{r_0} ) = 1 $
is subject to light rays from an omni-directional light (light source emitting light rays in all directions). The light source is located at $\mathbf{p_0}$. I'd like to find a closed form expression for the area of the shadow of this ellipse cast on the the plane $\mathbf{n}^T (\mathbf{r} - \mathbf{r_1} ) = 0 $. It is assumed that the ellipsoid lies between the light source and the plane.
My approach is detailed in my answer below.
Your comments, or alternative solutions, are highly appreciated.
First, I'll outline the method I used to find the closed form expression of the area of the shadow.
Step 1: Transform the ellipsoid into a unit sphere centered at the origin. The cone of light becomes a right circular cone.
Step 2: Find the area of the shadow of the shadow in the transformed coordinates by intersecting the transformed plane with the right circular cone resulting from the transformation.
Step 3: From the area found in Step 2, and the distance between the transformed light source location and the transformed plane, calculate the volume of the cut right circular cone.
Step 4: Calculate the volume of the original elliptical cone of shadow cast by the ellipsoid, by reversing the effect of the transformation.
Step 5: Calculate the distance between the light source and the given plane, and from the formula for the volume of a cone (which is now known, from Step 4), calculate the area of the base.
I will not put all the details of the above steps. However, the final result is as follows:
Define $\mathbf{v} = \mathbf{p_0} - \mathbf{r_0} $ and $\mathbf{ w }= \mathbf{p_0} - \mathbf{r_1} $
Then the area of the shadow is given by
$ A_{\text{shadow}} = \pi \dfrac{( \mathbf{n}^T \mathbf{w} )^2 \sqrt{ \mathbf{v}^T Q \mathbf{v} - 1 }} { \sqrt{\det(Q)} ( (\mathbf{n}^T \mathbf{v} )^2 - \mathbf{n}^T Q^{-1} \mathbf{n} )^{(3/2)}} $