Suppose you're given the paraboloid
$ z = a x^2 + b y^2 + c $
which you're viewing from the point $A$ that lies outside. What will be the equation of the cone of view of the paraboloid from $A$?
My approach:
The matrix-vector form of the paraboloid is obtained by defining $ r= [x, y, z]^T $, and
$ Q = \begin{bmatrix} a && 0 && 0 \\ 0 && b && 0 \\ 0 && 0 && 0 \end{bmatrix} $ and $q = \begin{bmatrix} 0 \\ 0 \\ - 1\end{bmatrix} $, then the equation is
$ r^T Q r + q^T r + c = 0 $
A ray originating at $A$ and having a direction vector $V$ is given parametrically by
$ r = A + t V $
Substitute this into the equation of the paraboloid
$ (A + t V)^T Q (A + t V) + q^T (A + t V) + c = 0 $
expand
$ t^2 (V^T Q V) + t ( 2 V^T Q A + q^T V) + A^T Q A + q^T A + c = 0 $
Since we want this ray to be tangential to the paraboloid, the two have to intersect in a signle point, which is only possible if the discriminant of the above quadratic equation in $t$ is zero. Hence,
$ (2 V^T Q A + q^T V)^2 - 4(V^T Q V) (A^T Q A + q^T A + c ) = 0 $
which can be written as the following quadratic form
$ V^T \left( (2 Q A + q )(2 Q A + q)^T - 4(A^T Q A + q^T A + c) Q \right) V = 0 $
Since $V = \dfrac{1}{t} ( r - A ) $
Then the equation of the cone of view of the paraboloid is
$ (r - A)^T Q_c (r - A) = 0$
where
$ Q_c = (2 Q A + q )(2 Q A + q)^T - 4(A^T Q A + q^T A + c) Q $
I've implemented this and graphed it on this SAGE page
It looks right. The plot also features the tangency curve which is an ellipse shown in blue.
Your comments, or alternative solutions are highly appreciated.
Thank you all.