Determine equation of plane that intersects an ellipsoid which results in a specified eccentricity of the resulting intersection ellipse

Question

Suppose you're given the ellipse

$$ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1 $$

I would like to find the direction of the normal to the cutting plane to this ellipsoid that will result in an ellipse with a given eccentricity.

That is, if the cutting plane is

$$ n^T (x - x_0) = 0 $$

where $n$ is a unit vector, then find $n$ such that the ellipse of intersection has eccentricity $e$, $0 \le e \lt 1 $.

Since the eccentricity of the intersection ellipse depends only on the direction of the normal vector of the cutting plane, and not its position, we can take $x_0 = 0 $.

My attempt:

Let the unit normal vector of the cutting plane be given in spherical coordinates as

$ n = [ \sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta ]^T $

Then it follows that two mutually perpendicular vectors that span the plane $n^T r = 0 $ are

$u_1 = [ \cos \theta \cos \phi, \cos \theta \sin \phi, - \sin \theta ]^T$

and

$u_2 =[- \sin \phi, \cos \phi, 0 ]^T $

Therefore, by defining $V = [u_1, u_2] $, we can express points of the plane as

$ r = V w$

where $w = [w_1, w_2]^T \in \mathbb{R}^2 $

Let $Q = \text{diag}( \dfrac{1}{a^2}, \dfrac{1}{b^2} , \dfrac{1}{c^2} ) $

Substituting $r$ in the equation of the ellipsoid, results in

$ w^T V^T Q V w = 1 $

The matrix $V^T Q V$ is $2 \times 2$, and by finding its eigenvalues $\lambda_1 , \lambda_2$ where $\lambda_1 \le \lambda_2 $, we have that the ratio of the semi-minor axis to the semi-major axis is

$\dfrac{B}{A} = \sqrt{ \dfrac{ \lambda_1 }{\lambda_2 } }$

So that the eccentricity of the ellipse of intersection is

$ e = \sqrt{ 1 - \left(\dfrac{B}{A}\right)^2 } = \sqrt{1 - \dfrac{\lambda_1}{\lambda_2} } $

So given $e$ we want to have the ratio of eigenvalues equal to

$ \dfrac{\lambda_1}{\lambda_2} = 1 - e^2 $

Since the normal of the plane has two degrees of freedom, namely $\theta$ and $\phi$, while the condition on the plane is one-dimensional, then we can fix one of the two parameters, for example $\phi$, and find the angle $\theta$ using iterative numerical methods (because the expression for the eigenvalues is very complex).

That's sums up my approach.

I appreciate any hints, comments, or solutions.

Answer 1

As mentioned above in the question, we'll take the unit normal to the plane to be

$ n = [\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta ]^T $

Then the $V$ matrix whose two columns are two unit vectors that are mutually orthogonal and also orthogonal to $n$ is given by

$ V = \begin{bmatrix} \cos \theta \cos \phi && - \sin \phi \\ \cos \theta \sin \phi && \cos \phi \\ -\sin \theta && 0 \end{bmatrix} $

It follows that

$V^T Q V = \begin{bmatrix} c_1^2 (A c_2^2 + B s_2^2 ) + C s_1^2 && (B - A) c_1 c_2 s_2 \\ (B - A) c_1 c_2 s_2 && A s_2^2 + B c_2^2 \end{bmatrix} $

where $c_1 = \cos \theta, s_1 = \sin \theta, c_2 = \cos \phi, s_2 = \sin \phi $, and $A = \dfrac{1}{a^2}, B = \dfrac{1}{b^2} , C = \dfrac{1}{c^2} $

Suppose now that we fix $\phi$ to a certain value, then we can define the following constants to simplify the analysis

$A_1 = A c_2^2 + B s_2^2 , A_2 = A s_2^2 + B c_2^2 , A_3 = (B-A) c_2 s_2$

Then our matrix becomes

$ V^T Q V = \begin{bmatrix} A_1 c_1^2 + C s_1^2 && A_3 c_1 \\ A_3 c_1 && A_2 \end{bmatrix} $

Its eigenvalues are the roots of the following characteristic polynomial,

$ | \lambda I - V^T Q V | = (\lambda - (A_1 c_1^2 + C s_1^2) ) (\lambda - A_2) - A_3^2 c_1^2 $

So the two $\lambda$'s are given by

$\lambda_1 = \dfrac{1}{2} ( B_1 - \sqrt{\Delta} ) $

$\lambda_2 = \dfrac{1}{2} ( B_1 + \sqrt{\Delta} ) $

where

$ B_1 = A_1 c_1^2 + C s_1^2 + A_2 $

$ \Delta = (A_1 c_1^2 + C s_1^2 + A_2)^2 - 4 ( A_2 (A_1 c_1^2 + C s_1^2) - A_3^2 c_1^2) $

Now we want

$\dfrac{\lambda_1}{\lambda_2} =1- e^2 = r^2 $

Therefore,

$ r^2 \lambda_2 = \lambda_1 $

Hence,

$ r^2 ( B_1 + \sqrt{\Delta} ) = ( B_1 - \sqrt{\Delta} ) $

Re-arranging,

$ \sqrt{\Delta} (r^2 + 1) = (B_1)(1 - r^2) $

Squaring

$ \Delta = K B_1^2 \hspace{35pt}(*) $

where $ K = \left( \dfrac{1 - r^2}{1 + r^2} \right)^2$

Recall that

$ B_1 = A_1 c_1^2 + C s_1^2 + A_2 =\alpha_1 + \alpha_2 \cos(2 \theta) $

where

$\alpha_1 = A_2 + \dfrac{1}{2} (A_1 + C) $

$\alpha_2 = \dfrac{1}{2} (A_1 - C) $

and

$ \Delta = (A_1 c_1^2 + C s_1^2 + A_2)^2 - 4 ( A_2 (A_1 c_1^2 + C s_1^2) - A_3^2 c_1^2) =\alpha_3 + \alpha_4 \cos(2 \theta) + \alpha_5 \cos^2(2 \theta)$ $

where

$\alpha_3 = \bigg(A_2 + \dfrac{1}{2} (A_1 + C) \bigg)^2 -2 A_2 (A_1 + C) + 2 A_3^2 $

$\alpha_4 = A_2 (A_1 - C) - 2 ( A_2 A_1 - A_3^2 - C ) $

$\alpha_5 = \dfrac{1}{4} (A_1 - C)^2$

Equation $(*)$ now becomes

$\alpha_3 + \alpha_4 \cos(2 \theta) + \alpha_5 \cos^2(2 \theta) = K (\alpha_1 + \alpha_2 \cos(2 \theta) )^2 $

which can be re-written in compact form as

$ \alpha \cos(2 \theta)^2 + \beta \cos(2 \theta) + \gamma = 0 \hspace{35pt} (**)$

where

$ \alpha = \alpha_5 - K\alpha_2^2 $

$ \beta = \alpha_4 - 2 K\alpha_1 \alpha_2 $

$ \gamma = \alpha_3 - K\alpha_1^2 $

And this is quadratic polynomial in $ \cos(2 \theta) $, which can be solved by using the quadratic formula.

As an example, I let $a = 3, b = 5, c = 4$, and plotted the resulting $\theta$ as a function of $\phi$ for an eccentricity of $ e = 0.7 $ over the range $0^\circ \le \phi \le 90^\circ$.