Identifying a paraboloid tangent to a number of planes

Question

Context: Recently, I investigated the problem of finding a parabola tangent to $4$ given lines in the Cartesian plane. I was able to write a program that finds this parabola. I wanted to extend this problem to $3D$. I figured from the dimensional analysis of the variables of the problem, that $7$ planes are needed to specify the circular paraboloid to be identified. However I am running in technical difficulties, and the iterative methods I've developed aren't converging.

Question: First, is it true that $7$ planes are needed to specify a circular paraboloid that is tangent to them ? Second, what methods can be used to identify such a paraboloid?

Answer 1

Hint about the methodology (not a full answer, due to lack of time).

Usually, this kind of issues are managed by using duality.

Here is a full treatment for the 2D case ; up to you for extending it to the 3D case.

Starting from a conic curve with equation $ax^2+2bxy+... =0$ associated with matrix

$$C = \begin{bmatrix}a&b&d\\b&c&e\\d&e&f\end{bmatrix}\tag{1}$$

we associate to it its dual conic, represented by its adjugate matrix :

$$C_a = \begin{bmatrix} cf-e^2 & de-bf & be-cd \\ de-bf & af-d^2 & bd-ae \\ be-cd & bd-ae & ac-b^2\end{bmatrix}\tag{2}$$

(when $A$ is invertible, $C_a$ is plainly $\det(A)A^{-1}.$)

$C_a$ is the matrix associated with dual conic with equation :

$$(cf-e^2)\lambda^2+2(de-bf)\lambda \mu +...=0,\tag{3}$$

(3) being a necessary and sufficient condition for a line with (homogeneous) equation

$$\lambda x + \mu y + \nu=0\tag{4}$$

to be tangent to the given conic curve.

See excellent explanations in this answer to which I have borrowed its notations.

I have a personal preference to represent (3) under the more compact form of $4 \times 4$ equivalent determinantal equation :

$$\begin{vmatrix} 0 & \lambda & \mu & \nu \\ \lambda & a & b & d \\ \mu & b & c & e\\ \nu & d & e & f\end{vmatrix}=0\tag{5}$$

The conic curve is a parabola if and only if this curve has a single point at infinity, a condition that is expressed by the fact that the lower right entry of $C_a$ is zero ($ac-b^2=0$).

For example, if we consider parabola with equation $-x^2+y=0$ to which are associated the following coefficients

$$a=-1,e=\tfrac12,b=c=d=f=0$$, we get the following tangential equation :

$$\tfrac14\lambda^2-\mu\nu=0\tag{6}$$

Giving arbitrary values to $\lambda$ and $\mu$ one gets a value of $\nu$ ; it remains to plot the line with equation (4) with these values.

The 3D case will have to manage the same kind of issues with

$$\begin{vmatrix} 0 & \lambda & \mu & \nu & \xi\\ \lambda & a & b & d & g\\ \mu & b & c & e & h\\ \nu & d & e & f & i \\ \xi & g & h & i & j\end{vmatrix}=0\tag{5}$$

Answer 2

The algebraic (implicit) equation of a circular paraboloid is

$ (r - V)^T R D R^T (r - V) + b_0^T R^T (r - V) = 0 $

where $r = [x, y, z]^T $ is a point on the paraboloid surface, and $V = [V_x, V_y, V_z]^T$ is the coordinate of the vertex. $b_0 = [0, 0, -1]$ and $R$ is a $3 \times 3 $ rotation matrix whose third column points in the direction of the axis of the circular paraboloid along the opening direction. Diagonal matrix $D$ is given by

$ D = \begin{bmatrix} a && 0 && 0 \\ 0 && a && 0 \\ 0 && 0 && 0 \end{bmatrix} $

Since this is a circular paraboloid, the rotation matrix $R$ is specified by only $2$ parameters, which give the direction of the third column of $R$. Adding the $3$ coordinates of the vertex and $a$, we have a total of $6$ parameters needed to fully specify the circular paraboloid. Hence from the dimensional point of view, we will need $6$ tangent planes to determine the paraboloid.

Define the $4 \times 4$ matrix

$ Q = \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 - b0^T R^T V \end{bmatrix} $

Then the equation of the circular paraboloid is

$ X^T Q X = 0 $

where $X = \begin{bmatrix} r \\ 1 \end{bmatrix} $

If we define the tangent plane by $N^T X = 0 $, then we must have

$ N_i^T Q^{-1} N_i = 0 \hspace{25pt}(***)$

for each of the six tangent planes.

So what I did, was to create the unknowns vector

$ Y = [ V_x, V_y, V_z, a , \theta, \phi ] $

From which, at any given of $Y$, I can compute matrix $R$, $D$, and therefore, $Q$. Inverting $Q$ and using $(***)$ for each of the six planes, I compute the "output" vector $Z$ where $Z_i = N_i^T Q^{-1} N_i $.

Now I run the unknown vector $Y$ through the Newton-Raphson multivariate root finder so that all the $Z_i \to 0$.

I implemented this with a small piece of code, where I created a sample circular paraboloid, and $6$ arbitrary planes, and an arbitrary initial guess of the parameter vector, and indeed the Newton-Raphson recursive algorithm converged to the true parameter vector of the set paraboloid.