Finding dual conic

Question

How do you find the dual conic associated with a conic and also a degenerated conic in matrix form? I have been attempting to find the intersection of two conics and the dual conic is a key step which I am having trouble figuring out.

Answer 1

For starters, if a nonsingular conic is given by $x^\top\!Ax=0$ for a nonsingular symmetric matrix $A$, then the dual conic is given by $x^\top\!A^{-1}x=0$.

Answer 2

This is really going an exercise in projective geometry. The book "Perspectives on Projective Geometry" by Richter-Gebert is a great resource. It can be hard to understand with self-study, but it's definitely a rich and valuable text to understanding conics.

A primal conic is described as the set of all points that satisfies a quadratic form $p^TAp=0$, where $p$ is in homogeneous coordinates and $A$ is a $3\times3$ matrix. For potentially every nonsymmetric matrix $A$, the matrix $(A + A^T)/2$ satisfies the quadratic form for the same set of points. The text above refers to this process as symmetrization. Since we can symmetrize when needed, we will just assume $A$ is symmetric from here.

If the matrix $A$ is nondegenerate (i.e. $\det(A) \neq 0$), a bijective mapping $p \rightarrow Ap$ can be created. If we consider that $Ap$ describes a line, $p^TAp$ can be seen as $\left \langle p, l\right \rangle = \left \langle p, Ap\right \rangle = p^TAp = 0$. This line is called a polar and the point $p$ is called a pole. Notice also that when the quadratic form is satisfied, $p$ is on the line $Ap$ and thus $Ap$ is tangent to the primal conic.

Taking the quadratic form $p^TAp = 0$ and remembering $A$ is symmetric, we can also transform it: $$\begin{align}p^T(AA^{-1})Ap &= 0 \\ p^TA^TA^{-1}Ap &= 0 \\ (Ap)^TA^{-1}(Ap) &= 0\end{align}$$ This modified form can be interpreted as the dual of the primal conic described by the quadratic form $l^TA^{-1}l=0$, where every $l=Ap$ that satisfies this equation is tangent to the primal conic.

Additionally, the adjugate of $A$ can mostly be substituted for the inverse because $A^{-1}=\operatorname{adj}(A)\,/\,\det(A)$.