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.


1 Answer 1


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.

  • $\begingroup$ Yes, that all sounds reasonable -- but can you construct it? $\endgroup$ May 7, 2015 at 3:38
  • $\begingroup$ @Mike: Since a ruler and compass construction can usually construct arbitrary many points on a resulting conic, but never the conic itself, when do you consider a conic as being constructed? Theoretically it's certainly possible to turn all the arithemtic needed above into constructions, but there should be something more elegant. Don't know yet. $\endgroup$
    – MvG
    May 7, 2015 at 4:16
  • 1
    $\begingroup$ I guess what I mean is to construct any of the "standard" ways in which a conic is determined, e.g. five points on it, or a projective polarity, or a non-perspective projectivity between two pencils, etc. $\endgroup$ May 8, 2015 at 4:25

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