Tangent Planes to an ellipsoid

The points $r$ on the ellipsoid satisfy the equation

$r^t A r = 1$

where diagonal matrix A is given by:

$A = \begin{bmatrix} 1/a^2 & 0 & 0 \\ 0 & 1/b^2 & 0 \\ 0 & 0 & 1 / c^2 \end{bmatrix}$

The normal to the surface of the ellipsoid, is the gradient, given by:

$g = 2 A r$

Now, the plane tangent to the ellipsoid at point $r_0$ has the equation:

$(2 A r0) \dot (r - r0) = 0$

hence,

${r_0}^t A (r - r_0) = {r_0}^t A r - {r_0}^t A r_0 = 0$

from which

${r_0}^t A r = 1$

Since the tangent plane passes through $p_0$ and $p_1$ , then

${r_0}^t A p_0 = 1$ or $(A p_0)^t r_0 = 1$

and

${r_0}^t A p_1 = 1$ or $(A p_1)^t r_0 = 1$

These two equations are equations of planes. Since $r_0$ satisfies both of them, it lies on their intersection, which is a line.

Hence, solving the linear system of the two equations after substituting the numerical values of $A$ , $p_0$ and $p_1$

results in,

$r_0 = q_0 + t d$

where $q_0$ is a fixed point, and $d$ is the direction vector of the line, and $t$ is a free parameter.

Finally, we substitute this form of $r_0$ into the equation of the ellipsoid,

$(q_0 + t d)^t A (q_0 + t d) = 1$

This simplifies to

$a_2 t^2 + a_1 t + a_0 = 0$

where

$a_2 = d^t A d$

$a_1 = 2 {q_0}^t A d$

$a_0 = {q_0}^t A q_0 - 1$

Next we solve this quadratic equation using the quadratic formula to obtain the two roots $t_1$ and $t_2$ , corresponding to which, we have

$r_1 = q_0 + t_1 d$ and $r_2 = q_0 + t_2 d$

And it follows that the gradients at these two points are given by:

$g_1 = 2 A r_1$ and $g_2 = 2 A r_2$

At the very final step, we find the angle between the two gradients (normals to the surface), and substract this angle from 180, to obtain the angle between the two planes.

Using simple computer code, this results in the angle between the two planes being $\boxed{75.6 }$