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

Question

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.

Answer 1

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 $c_3$.

Answer 2

Thanks intelligenti-pauca for pointing me in the right direction! I'll add some details on my approach here.

Let $\mathbf{P} = \begin{bmatrix} \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{a} \end{bmatrix}$ be the matrix of eigenvectors of $\mathbf{Q}_r$. I want to compose $\mathbf{c}$ in the eigenbasis of $\mathbf{Q}_r$ as $\mathbf{c} = \mathbf{P} \mathbf{d}$, so I need to find the elements $d_1$, $d_2$, $d_3$ of $\mathbf{d}$ as intelligenti-pauca suggested such that

$$\mathbf{c} = \mathbf{v}_1 d_1 + \mathbf{v}_2 d_2 + \mathbf{a} d_3 = \mathbf{P}\mathbf{d}$$

I find the first two elements $d_1$ and $d_2$ by an inner product with $\mathbf{q}_d$:

$$d_1 = \mathbf{v}_1^T\mathbf{q}_d = \mathbf{v}_1^T\mathbf{a}\frac{b^2}{2} - \mathbf{v}_1^T\mathbf{Q}_r\mathbf{c} = - \mathbf{v}_1^T\mathbf{Q}_r\mathbf{c}$$

The same holds for $d_2 = \mathbf{v}_2^T\mathbf{q}_d$, recall $\mathbf{v}_1^T\mathbf{a} = \mathbf{v}_2^T\mathbf{a} = 0$.

Now, to find $d_3$ I substitute $\mathbf{c} = \mathbf{P}\mathbf{d}$ into the expression for $q_0$:

$$\begin{aligned} q_0 &= -b^2\mathbf{a}^T\mathbf{c} + \mathbf{c}^T\mathbf{Q}\mathbf{c} \\ &= -b^2\mathbf{a}^T \mathbf{P} \mathbf{d} + \mathbf{d}^T\mathbf{P}^T\mathbf{Q}_r\mathbf{P}\mathbf{d} \\ &= -b^2\begin{bmatrix} 0 & 0 & 1 \end{bmatrix}\mathbf{d} + \mathbf{d}^T \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\mathbf{d} \\ &= -b^2d_3 - d_1^2 - d_2^2 \\ &\implies d_3 = \frac{q_0 + d_1^2 + d_2^2}{-b^2} \end{aligned}$$

Now that I have $d_1$, $d_2$, and $d_3$ I transform them back into the original space to get $\mathbf{c}$: $$\mathbf{c} = \mathbf{P}\mathbf{d}$$