Axis of an oblique elliptical cone

The base is the given ellipse, whose edge can be written parametrically as $b(t) = v_0 + v_1 \cos t + v_2 \sin t$ , where $v_0 = (0, 0, 0) , v_1 = (10, 0, 0) , v_2 =(0, 6, 0)$ . The surface of the cone is parameterized as follows: If $p$ is on the surface of the cone, then it is on the line segment joining a point on the base with apex. Therefore, there exists $s \in [0, 1]$ , and $t \in [0, 2\pi]$ , such that,

$p = A + s ( b(t) - A ) \hspace{8pt} (1)$

Plugging the expression for $b(t)$ ,

$p = A + s (v_0 - A + v_1 \cos t + v_2 \sin t ) \hspace{8pt} (2)$

Hence,

$p - A = s M v(t) \hspace{8pt} (3)$

where,

$v(t) = \begin{bmatrix} \cos t \\ \sin t \\ 1 \end{bmatrix}\hspace{8pt} (4)$

and

$M = \begin{bmatrix} v_1 && v_2 && (v_0 - A ) \end{bmatrix} \hspace{8pt} (5)$

We note that the vector $v(t)$ satisfies the quadratic form equation $v(t)^T Q_0 v(t) = 0$ where,

$Q_0 = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && -1 \end{bmatrix}\hspace{8pt} (6)$

And from equation (3), we have

$s v(t) = M^{-1} ( p - A ) \hspace{8pt} (7)$

Hence,

$s^2 v(t)^T Q_0 v (t) = 0 = (p - A)^T M^{-T} Q_0 M^{-1} (p - A) \hspace{8pt} (8)$

So the equation of the cone is,

$( p - A)^T Q ( p - A) = 0 \hspace{8pt} (9)$

where

$Q = M^{-T} Q_0 M^{-1} \hspace{8pt} (10)$

Since $Q$ is symmetric, its eigenvalues are real, and since $Q_0$ has two positive eigenvalues and one negative eigenvalue, then so will the matrix $Q$ .

The axis of the cone is the eigenvector corresponding to the negative eigenvalue. This can be seen by diagonalizing Q into,

$Q = R D R^T \hspace{8pt} (11)$

Define the vector $w = R^T (p - A)$ , then equation (9) becomes,

$w^T D w = 0 \hspace{8pt} (12)$

If $w = [w_1, w_2, w_3]$ , then equation (12) reads,

$D_{11} w_1^2 + D_{22} w_2^2 + D_{33} w_3^2 = 0 \hspace{8pt} (13)$

And suppose that $D_{33}$ is negative, then (13) is the standard equation of an elliptical cone in the $w$ -frame. Its axis is the $w_3$ axis, and from this it follows that the axis of our cone is along the third column of the eigenvectors matrix $R$ .

Assuming we have a computer program to find the eigenvalues and eigenvectors of a $3 \times 3$ symmetric matrix, then our problem is almost solved now that we have the direction vector $d$ of the axis. What remains is to write the equation of the line passing through the apex and having this direction vector, in parametric form, it is,

$q(t) = A + t d \hspace{8pt} (14)$

Since we want the intersection with $xy$ plane, we set the $z$ coordinate of $A + t d$ to zero, and this gives $t = \dfrac{-A_z}{d_z}$ .

Hence, $I_x = A_x + t d_x$ and $I_y = A_y + t d_y$ . The axis unit direction vector is $(0.4769513, 0.580052181, 0.660346064)$ , and the intersection point is $(2.777252316, 1.215936417, 0 )$ . The answer is $I_x + I_y \approx 3.993$