Can the vertex of an elliptic paraboloid be recovered from its quadric matrix form

conic sectionsgeometryquadratic-formsquadrics

A circular paraboloid with unit axis $\mathbf{a}$, vertex $\mathbf{c}\in\mathbb{R}^3$, and quadratic coefficient $b\in\mathbb{R}$ is defined implicitly as the set of points $\mathbf{r}\in\mathbb{R}^3$ such that the following equality holds:

$$
\mathbf{a}^T(\mathbf{r} – \mathbf{c})\big(b^2 + \mathbf{a}^T(\mathbf{r} – \mathbf{c})\big) + ||\mathbf{r} – \mathbf{c}||^2 = 0
$$

I can identically write this as a quadric form in homogeneous coordinate $\mathbf{r}_h = \begin{bmatrix} \mathbf{r} \\ 1\end{bmatrix}$ as

$$
\mathbf{r}_h^T\mathbf{Q}\mathbf{r}_h = 0
$$

where $\mathbf{Q}$ is symmetric indefinite. $\mathbf{Q}$ can be composed of the original parameters $\mathbf{a}$, $\mathbf{c}$, and $b$ as

$$
\mathbf{Q} = \begin{bmatrix} \mathbf{Q}_r & \mathbf{q}_d \\ \mathbf{q}_d^T & q_0 \end{bmatrix}
$$

where $\mathbf{Q}_r = \mathbf{a}\mathbf{a}^T – \mathbf{I}$, $\mathbf{q}_d=\frac{b^2}{2}\mathbf{a} – \mathbf{Q}_r\mathbf{c}$, and $q_0 = -b^2\mathbf{a}^T\mathbf{c} + \mathbf{c}^T\mathbf{Q}_r\mathbf{c}$.

Now, let's say I'm given the matrix $\mathbf{Q}$. I want a procedure for recovering the original parameters $\mathbf{a}$, $\mathbf{c}$, and $b$ from $\mathbf{Q}$. Here are my steps:

  1. Let $\lambda_{max}$ be the maximum eigenvalue of $\mathbf{a}\mathbf{a}^T = \mathbf{Q}_r + \mathbf{I}$, and $\mathbf{v}_{max}$ be the corresponding eigenvector. I know that $\mathbf{v}_{max} = \mathbf{a}$.
  2. I know $\mathbf{a}\in\mathcal{N}(\mathbf{Q}_r)$ and $\mathbf{a}^T\mathbf{a}=1$ (and I recovered $\mathbf{a}$ in the last step), so it follows that $\mathbf{q}_d^T\mathbf{a} = \frac{b}{2} \implies b = 2\mathbf{q}_d^T\mathbf{a}$.
  3. Now I am stumped. I cannot find any tricks that will give me $\mathbf{c}$ given $\mathbf{Q}$, $\mathbf{a}$, and $b$.

I am wondering if it is possible to recover $\mathbf{c}$ at all from the matrix $\mathbf{Q}$.

I have tried using the normal vector to the surface: $\frac{\partial}{\partial \mathbf{r}_h} \mathbf{r}_h^T\mathbf{Q}\mathbf{r}_h\Bigg\vert_{\mathbf{r} = \mathbf{c}} = 2\mathbf{r}_h^T \mathbf{Q} \Bigg\vert_{\mathbf{r} = \mathbf{c}} = 2\mathbf{c}_h^T\mathbf{Q} = \mathbf{a}_h^T \implies \mathbf{c}_h=\frac{1}{2}\mathbf{Q}^{-1}\mathbf{a}_h$, which seems to be sensitive to the homogeneous coordinate I choose for $\mathbf{a}_h$.

I have read here that because $\det(\mathbf{Q}_r) = 0$ for all paraboloids, paraboloids are classed as "non-central" quadric surfaces, but I have not found any information about what that means for identifying $\mathbf{c}$ based on $\mathbf{Q}$.

Thank you for any suggestions!

Edit: I fixed a typo in definition of $q_0$ that didn't affect the overall problem.

Best Answer

I think your last try is the good one: the gradient must be proportional (not equal) to $\mathbf{a}$: $$ 2\mathbf{Q}_r\mathbf{r}+2\mathbf{q}_d = k\mathbf{a}. $$ Take the scalar product of both members with the other eigenvectors $\mathbf{v}_1$ and $\mathbf{v}_2$ of $\mathbf{Q}_r$ to find the projections $c_1$ and $c_2$ of $\mathbf{c}$ on them. You'll have then: $$ \mathbf{c}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{a}. $$ Finally, plug this value of $\mathbf{c}$ into the equation of the paraboloid to find $c_3$.

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