Paraboloid Vertex and Axis (Simplified)

Fine problem. I use Python. I find point $(x_M , y_M, z_M)$ on surface

$-1865 x^2 - 774 xy + 1220 xz + 5306x - 2853y^2 - 4368 y z - 10991y - 2125z^2 + 12128z + 718205=0$

with

$K(x_M,y_M,z_M)=Max( K(x,y,z))$ .

$K(x,y,z)$ is here GaussianCurvature - formula $(16)$ .

$x_M=-15.204$

$y_M= 24.896$

$z_M= -29.994$

$K=0.13645128106808901$

$Answer= 117.8718640777565$

The original paraboloid can be written in matrix-vector form as

$p^T Q p + a^T p + b = 0 \hspace{24pt} (1)$

where

$Q = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 0 \end{bmatrix}$

$a^T = \begin{bmatrix} 0, 0, -1 \end{bmatrix}$

$b = -5$

Now, since the coordinate vector of a point on the new paraboloid is $p' = A p$ , then $p = A^{-1} p'$ . Plugging this into the above equation results in

$p'^T A^{-T} Q A^{-1} p' + a^T A^{-1} p' + b = 0 \hspace{24pt} (2)$

This is the equation of the new paraboloid. To identify the vertex and axis, we have to put it in the standard form of a general paraboloid which is,

$(p - r_0)^T R D R^T (p - r_0) + a^T R^T (p - r_0) = 0 \hspace{24pt} (3)$

where $D$ is a diagonal matrix with $D_{11} > 0 , D_{22} > 0 , D_{33} = 0$ , $R$ is a rotation matrix, $r_0$ is the vertex, and $a$ is as above. The third column of matrix $R$ is along the axis of the paraboloid.

The standard equation of a general paraboloid can be multiplied through by a nonzero constant, and this results in

$\lambda (p - r_0)^T R D R^T (p - r_0) + \lambda a^T R^T (p - r_0) = 0 \hspace{24pt} (4)$

We want to find the value of $\lambda, D, R, r_0$ such that equations $(2) , (4)$ are identical, that is,

$A^{-T} Q A^{-1} = \lambda R D R^T \hspace{24pt} (5)$

$A^{-T} a = - 2 \lambda R D R^T r_0 + \lambda R a \hspace{24pt} (6)$

and

$b = \lambda r_0^T R D R^T r_0 - \lambda a^T R^T r_0 \hspace{24pt} (7)$

By diagonalizing $A^{-T} Q A^{-1}$ we immediately find $R$ and the diagonal matrix $\lambda D$ . Next, by pre-multiplying equation $(6)$

by $R^T$ it becomes

$R^T A^{-T} a = - 2 \lambda D R^T r_0 + \lambda a \hspace{24pt} (8)$

The third row of this vector equation reads,

$- (R^T A^{-T})_3 = - \lambda$

because the third row of matrix $D$ is a row of zeroes. Thus, we have found $\lambda$ , and therefore, matrix $D$ is now known.

Finally, we have to find the vertex $r_0$ . Plugging the values we have for $R , D , \lambda$ into equation $(6)$ , we can solve this system of equations for $r_0$ . The solution is of the form $r_0 = v_0 + t v_1$ where $t$ is yet to be determined. To determine it, we use equation $(7)$ . The term $r_0^T R D R^T r_0$ in equation $(7)$ , can be simplified by using equation $(6)$ , where we note that pre-multiplying by $r_0^T$ , and re-arranging, results in,

$\lambda r_0^T R D R^T r_0 = \dfrac{1}{2} \left( \lambda r_0^T R a - r_0^T A^{-T} a \right)$

Plugging this into equation $(7)$ results in

$2 b = (- \lambda a^T R^T - a^T A^{-1} ) r_0 \hspace{24pt} (9)$

Equation $(9)$ , coupled with $r_0 = v_0 + t v_1$ can be used directly to find the constant $t$ . Hence $r_0$ is now determined.

That was an outline of the solution method for this problem. For the numerical results, a small computer program can be used, provided you have a routine for finding the eigen structure of a symmetric matrix.