# Recovering a conic from a pole-polar pair

Consider a conic section $C$ in $\mathbb{R}^2$. Every point $P$ in the plane has a "dual" (pole-polar duality) line $L$ with respect to $C$ such that lines $PA$ and $PB$ are tangent to $C$, where $L \cap C = \{A, B\}$. However, given $P, A, B$, is it possible to recover $C$? Is $C$ even uniquely determined by those three points?

My vain attempt is as follows. Suppose $C$ has the matrix representation $\mathbf{C}$, and the images of $P, A, B$ under the map $(x, y) \mapsto \begin{bmatrix}x & y & 1\end{bmatrix}^\intercal$ are $\mathbf{p}, \mathbf{a}, \mathbf{b}$ respectively. Since $A, B \in C$, $\mathbf{a}^\intercal\mathbf{C}\mathbf{a} = 0$ and $\mathbf{b}^\intercal\mathbf{C}\mathbf{b} = 0$. That is, $\mathbf{C}\mathbf{a}$ is orthogonal to $\mathbf{a}$ (note that $(\mathbf{a}^\intercal\mathbf{C})^\intercal = \mathbf{C}\mathbf{a}$ because $\mathbf{C}$ is symmetric). Similarly, $\mathbf{C}\mathbf{b}$ is orthogonal to $\mathbf{b}$.

From tangency we also have $\mathbf{p}^\intercal\mathbf{C}\mathbf{a} = 0$ and $\mathbf{p}^\intercal\mathbf{C}\mathbf{b} = 0$. Since scaling $\mathbf{C}$ doesn't change the fact that it represents $C$, we may set $\mathbf{C}\mathbf{p} = \mathbf{a} \times \mathbf{b}$.

Is it possible to proceed further than this?