Constructing a conic from two point-polar pairs Suppose I have two points $A$ and $B$ and two lines $a$ and $b$ in the (projective) plane.  Can I construct a conic section for which $a$ is the polar of $A$ and $b$ is the polar of $B$?  How unique or non-unique is it?
I know that if $A$ lies on $a$ and $B$ lies on $b$, so that the conic must be tangent to $a$ at $A$ and to $b$ at $B$, then it is uniquely determined by specifying a third point $C$ lying on it.  A construction in this case can be found for instance in $\S$8.4 of Coxeter's Projective geometry.
 A: 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.
