Given an elliptic paraboloid whose equation is
$ (r - V)^T R D R^T (r - V) + b_0^T R^T (r - V) = 0 $
where $r$ is a point on the surface of the paraboloid, and $V$ is its vertex. $R$ is a rotation matrix, and $D = \operatorname{diag}\{ a , b, 0 \} , a \gt 0, b \gt 0$. $b_0 = [0, 0, -1]^T $.
And also given a plane $n^T (r - r_1) = 0 $, I want to find the distance between the plane and the paraboloid.
My approach:
The equation of the paraboloid can be re-written as follows
$ X^T G X = 0 $
where $ X = \begin{bmatrix} r \\ 1 \end{bmatrix} $ and
$ G = \begin{bmatrix} R D R^T && - R D R^T V + \frac{1}{2} R b_0 \\ - V^T R D R^T + \frac{1}{2} b_0^T R^T && V^T R D R^T V - V^T R b_0 \end{bmatrix} $
Now I want to find the plane that is parallel to the given plane but tangent to the paraboloid. Such a plane would have the equation $N^T X = 0 $ where
$ N = \begin{bmatrix} n \\ d \end{bmatrix} $
where $n$ is the normal vector of the given plane, and $d$ is to be found. The condition for tangency is
$ N^T G^{-1} N = 0 $
Now,
$ G^{-1} = \begin{bmatrix} G_{11} && G_{12} \\ G_{12}^T && G_{22} \end{bmatrix} $
where $G_{11}$ is $ 3 \times 3 $ and $G_{12}$ is $3 \times 1$ and $G_{22}$ is $1 \times 1$ (a scalar).
It is easy to show (using cofactors) that $G_{22} = 0 $. Multiplying out the expression above, we get
$ n^T G_{11} n + 2 n^T G_{12} d = 0 $
and this can be solved for $d$, namely
$ d = - \dfrac{n^T G_{11} n}{ 2 n^T G_{12} }$
Now the original plane we're given has the equation $ n^T r + d_1 = 0 $
while the parallel plane that is tangent to the paraboloid has the equation $n^T r + d = 0 $
Hence, the distance between the given plane and the paraboloid is the distance between the two parallel planes which is given by
$ D_{min} = \dfrac{ | d - d_1 | } { \| n \| } $
Alternative methods, comments, hints, and solutions to this question are most welcome and appreciated.