How to find the tangent cone to an ellipsoid

Question

Is there a standard way to find the tangent cone to a general ellipsoid from a point outside of the ellipsoid?

Answer 1

$\def\x{\mathbf{x}}$ $\def\u{\mathbf{u}}$

Special case: For conics whose center is at the origin see Armstrong, Where is the cone?, Section 4.

Specifically, if the ellipsoid is defined as

$$\x^T A\x =1,$$

the paper shows that the tangent cone from $\u$ to the surface $\x^TA\x=1$ has the equation $$ \begin{equation*} \boxed{(\x^TA\u-1)^2=(\x^TA\x-1)(\u^TA\u-1).} \end{equation*} $$

General case: If the conic is not centered at the origin, see Bobenko et al, pg 17, Def 3.1.

For a symmetric matrix $Q$, and $\x,\u$ in homogeneous form, the tangent cone from $\u$ to the quadric $\x^TQ\x=0$ is given by

$$ (\x^TQ\u)^2-(\x^TQ\x)(\u^TQ\u)=0. $$

The special case uses a $3\times3$ matrix $A$, while the general case uses a $4\times4$ matrix $Q$, so arguably one could just use the general formula in all cases.