Locus of point from which three mutually perpendicular lines tangent to an ellipsoid can be drawn

Question

Inspired by this problem, I would like to find the locus of all points from which three mutually perpendicular tangent lines can be drawn to a given ellipsoid.

The ellipsoid is given by

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

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My attempt:

Let the vector $r = [x, y, z]^T$ be the position vector of a point on the ellipsoid. And let

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

Then the equation of the ellipsoid is

$$ r^T Q r = 1 $$

From a point $P$ outside the ellipsoid a line is constructed with a direction vector $V$, then the equation of this line is

$ r = P + t V $

This line is to intersect the ellipsoid in exactly one point. This way we know it is tangent to it. Substituting the equation of the line into the equation of the ellipsoid gives

$ (P + t V)^T Q (P + t V) = 1 $

Expand this you get

$ P^T Q P + 2 t V^T Q P + t^2 V^T Q V = 1 $

This is a quadratic equation in $t$ and will have exactly one root if and only if its discriminant is $0$. Therefore, we impose that to get

$ (V^T Q P)^2 - (P^T Q P - 1) (V^T Q V) = 0 $

This is actually the equation of all possible directions that satisfy tangency from P.

This is where I am at. I am not sure how to continue from here. Any hints or full solutions are highly appreciated.

Answer 1

Comment: In two d it seems that the locus of such points is a circle. Now if you revolve the construction about the major axis you get an ellipsoid and and sphere as the locus.This is when you work in front plane(yz plane) which is shown in picture a.

another case is when we draw the tangent on the right plane(xz plane) (picture b)and planes parallel xz plane. In this case the locus will be an ellipsoid.

The third case is when we draw tangents from space between the front and right planes(xyz space). In this case the locus of points can be a combination of sphere and ellipsoid or generally an ellipsoid.

The situation is shown in the third picture. I do not think finding the general equation representing all cases is easy.

Answer 2

Starting with the equation that I arrived at in my question

$ (V^T Q P)^2 - (P^T Q P - 1) (V^T Q V) = 0 $

This can be written as follows

$ V^T \bigg( Q P P^T Q - ( P^T Q P - 1) Q \bigg) V = 0 \tag{1}$

Let's first consider the two-dimensional case. Suppose $V$ and $W$ satisfy $(1)$ and that they're both unit vectors and they're perpendicular to each other.

Define the matrix

$ M = Q P P^T Q - ( P^T Q P - 1) Q \tag{2} $

So that

$ V^T M V = 0 \tag{3}$

and

$ W^T M W = 0 \tag{4}$

Suppose $V = (v_1, v_2) $ and $ W = (w_1, w_2) $, then the first equation says

$ M_{11} v_1^2 + M_{22} v_2^2 + 2 M_{12} v_1 v_2 = 0 \tag{5} $

Similarly the second equation says

$ M_{11} w_1^2 + M_{22} w_2^2 + 2 M_{12} w_1 w_2 = 0 \tag{6}$

Adding these two equations gives

$ M_{11} (v_1^2 + w_1^2) + M_{22} (v_2^2 + w_2^2) + 2 M_{12} (v_1 v_2 + w_1 w_2) = 0 \tag{7}$

Define the matrix $ R = [ V , W ] $, then

$ R R^T = I $ because $R$ is orthogonal. The $(1, 2)$ element of the left hand side is

$$ v_1 v_2 + w_1 w_2 $$

And this is equal to zero by comparing to the right hand side.

Also the $(1,1)$ entry of $R R^T$ is given by $ v_1^2 + w_1^2 $ and this is equal to $1$, and finally the $(2,2)$ entry of $R R^T$ is $v_2^2 + w_2^2$ which is also equal to $1$.

Therefore, equation $(7)$ becomes

$ M_{11} + M_{22} = 0 \tag{8}$

Recall that

$ M = Q P P^T Q - ( P^T Q P - 1) Q \tag{9} $

Equation $(8)$ says that we must $\text{Trace}(M) = 0 $, but

$ \text{Trace}(M) = \text{Trace}(Q P P^T Q) - ( P^T Q P - 1) \text{Trace}(Q) $

And this is equal to

$ \text{Trace}(M) = P^T Q^2 P - ( P^T Q P - 1) \text{Trace}(Q) $

Substituting $ P = [x, y]^T$ and $ Q = \begin{bmatrix} \dfrac{1}{a^2} && 0 \\ 0 && \dfrac{1}{b^2} \end{bmatrix} $, we end up with

$ \text{Trace}(M) = \dfrac{x^2}{a^4} + \dfrac{y^2}{b^4} - \left( \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - 1\right) \left( \dfrac{1}{a^2} + \dfrac{1}{b^2} \right) $

And this simplifies to

$ \text{Trace}(M) = - \left( \dfrac{1}{a^2 b^2} \right) (x^2 + y^2 ) + \dfrac{1}{a^2} + \dfrac{1}{b^2} = 0 $

Hence, the locus of $P$ is given by

$ x^2 + y^2 = a^2 + b^2 \tag{10}$

which is a circle.

I've worked the three-dimensional case in a similar fashion to the above, and the final result is that the locus of points $P = [x,y,z]^T $ is the ellipsoid

$ x^2 (b^2 + c^2 ) + y^2 (a^2 + c^2 ) + z^2 (a^2 + b^2) = a^2 b^2 + a^2 c^2 + b^2 c^2 \tag{11}$