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 want to find a closed form expression for the area of the shadow of the ellipsoid onto the plane $\mathbf{n}^T (\mathbf{r} - \mathbf{r_1} ) = 0 $.
My attempt: is presented in my answer below.
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.
For this, diagonalize $Q$ as $Q = R D R^T $, and define
$ \mathbf{r'} = D^{1/2} R^T (\mathbf{r - r_0}) $
Then
$ \mathbf{r'}^T \mathbf{r'} = 1 $
is the unit sphere centered at the origin.
Step 2: Find the area of the shadow in the transformed coordinates $r'$ by intersecting the transformed plane with the right circular cone resulting from the transformation.
The transformed plane equation is
$ \mathbf{n}^T ( \mathbf{r_0} + R D^{-1/2} \mathbf{r'} - \mathbf{r_1} ) = 0 $
i.e.
$ \mathbf{n'}^T ( \mathbf{r'} - D^{1/2} R ( \mathbf{r_1 - r_0} ) ) = 0 $
where $\mathbf{n'} = D^{-1/2} R^T \mathbf{n} $
The semi-vertical angle of the right circular cone is given by
$ \sin \theta = \dfrac{1}{\| \mathbf{p_0'} \| } $
where
$ \mathbf{p_0'} = D^{1/2} R^T (\mathbf{p_0 - r_0}) $
At this point, define $\mathbf{v} = \mathbf{p_0} - \mathbf{r_0} $ and $\mathbf{ w }= \mathbf{p_0} - \mathbf{r_1} $
$ \sin \theta = \dfrac{1}{\sqrt{ \mathbf{(p_0 - r_0)}^T Q \mathbf{(p_0 - r_0)} } } = \dfrac{1}{\sqrt{ \mathbf{v}^T Q \mathbf{v}} } $
The ellipse of the intersection between the (transormed) right circular cone of line and the transformed plane has it semi-axes as follows
$ a= \dfrac{ z_0 \tan \theta \cos \phi }{ (1 - \sin^2 \phi \sec^2 \theta) } $
$ b = \dfrac{ z_0 \tan \theta \cos \phi }{ (1 - \sin^2 \phi \sec^2 \theta )^{1/2} } $
where $\phi$ is the angle between the axis the right circular cone and the normal vector to the transformed plane $( \mathbf{n'} )$. These formulas are derived here. Therefore, the area is
$ A' = \pi a b = \pi \dfrac{ (z_0 \cos \phi)^2 \tan^2 \theta }{ (1 - \sin^2 \phi \sec^2 \theta)^{3/2}} $
Equivalently, this is equal to
$ A' = \pi \dfrac{ (z_0 \cos \phi)^2 \sin^2 \theta \cos \theta }{ (\cos^2 \theta - \sin^2 \phi)^{3/2}} = \pi \dfrac{ (z_0 \cos \phi)^2 \sin^2 \theta \cos \theta }{ (\cos^2 \phi - \sin^2 \theta )^{3/2} } $
It can be shown that
$z_0^2 = \left( \dfrac{\mathbf{n}^T \mathbf{w} }{ \mathbf{n}^T \mathbf{v} } \right)^2 \mathbf{v}^T Q \mathbf{v} $
and
$ z_0 \cos \phi = \dfrac{ | \mathbf{n}^T (\mathbf{p_0 - r_1}) |}{\sqrt{\mathbf{n}^T Q^{-1} \mathbf{n} }} = \dfrac{ |\mathbf{n}^T \mathbf{w} | }{\sqrt{\mathbf{n}^T Q^{-1} \mathbf{n}}} $
So that
$ \sec^2 \phi = \dfrac{ (\mathbf{v}^T Q \mathbf{v}) (\mathbf{n}^T Q^{-1} \mathbf{n}) }{ (\mathbf{n}^T \mathbf{v})^2 } $
Substituting these expressions and simplifying gives
$ A' = \pi \dfrac{ \sqrt{ \mathbf{n}^T Q^{-1} \mathbf{n}} (\mathbf{n}^T \mathbf{w} )^2 \sqrt{ \mathbf{v}^T Q \mathbf{v} - 1 } } { \left( (\mathbf{n}^T \mathbf{v})^2 - \mathbf{n}^T Q^{-1} \mathbf{n} \right)^{3/2} } $
Step 3: Now the volume of the right circular cone is
$ V' = \dfrac{1}{3} A' h' $ where $ h' = z_0 \cos \phi $
i.e.
$ V' = \dfrac{\pi}{3} \dfrac{ (\mathbf{n}^T \mathbf{w} )^3 \sqrt{ \mathbf{v}^T Q \mathbf{v} - 1 } } { \left( (\mathbf{n}^T \mathbf{v})^2 - \mathbf{n}^T Q^{-1} \mathbf{n} \right)^{3/2} } $
Step 4: The volume of the original cone is
$ V = \dfrac{V}{\sqrt{\det{Q}}} $
i.e.
$ V = \dfrac{\pi}{3} \dfrac{ (\mathbf{n}^T \mathbf{w} )^3 \sqrt{ \mathbf{v}^T Q \mathbf{v} - 1 } } {\sqrt{\det{Q}} \left( (\mathbf{n}^T \mathbf{v})^2 - \mathbf{n}^T Q^{-1} \mathbf{n} \right)^{3/2} } $
Step 5: To find the actual area of the shadow, use
$ A = \dfrac{3 V}{h}$
where $ h $ is the altitude and is given by $ h = \mathbf{n}^T \mathbf{w} $
This gives
$ 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)}} $