A conic is described by a symmetric $3\times3$ matrix, and multiples of a matrix describe the same conic. The polar line of a point can be obtained by multiplying that matrix with the point, but the result will only be unique up to scalar multiples. So you essentially have two vector equations:
\begin{align*}
\lambda a &= \begin{pmatrix}
m_{11} & m_{12} & m_{13} \\
m_{12} & m_{22} & m_{23} \\
m_{13} & m_{23} & m_{33}
\end{pmatrix} A &
\mu b &= \begin{pmatrix}
m_{11} & m_{12} & m_{13} \\
m_{12} & m_{22} & m_{23} \\
m_{13} & m_{23} & m_{33}
\end{pmatrix} B
\end{align*}
Spelled out component-wise these would be $6$ linear equations in $8$ variables. Since the situation is homogeneous, you can in general fix one of the variables arbitrarily, which leaves you with one real degree of freedom. So I'd expect a one-parameter family of possible solutions.
It should be possible to represent that freedom as a third point $C$ which has to lie on the conic as well, since that simply leads to a seventh equation. The requirement that these seven equations have to be linearily independent will give rise to some non-degeneracy conditions which I haven't worked out yet. For sufficiently general position everything should work fine.
