8

I recently had to compute the intersection of two conics, and found it to be a long and complicated procedure. Looking at Wikipedia, the advice to find the points of intersection of two conics in general, which is echoed in the answer to this question, is to write the conics in matrix form, say $C_1$ and $C_2$, and then solve $\det( \lambda C_1+\mu C_2)=0$ for values of $\lambda$ and $\mu$, then consider the degenerate conic $C_0=\lambda C_1+\mu C_2$ and the intersection of this with your original conics will be easier to compute (and coincides with the intersection of the two conics originally).

My question is: originally we were looking for the solutions of two degree 2 polynomials in two variables. How is solving $\det(\lambda C_1 + \mu C_2)=0$, a degree three polynomial in two variables, any easier (in general)? Certainly there are specific matrices $C_1$ and $C_2$ for which this is quite obvious, but in general it does not seem to me as though we have made our problem simpler.

Community
  • 1
mboratko
  • 4,553

1 Answers1

9

It's not "really" a polynomial in two variables: it's homogeneous in $\lambda$ and $\mu$, so really it behaves like a polynomial in one variable $\frac{\lambda}{\mu}$. The use of the homogeneous polynomial is that we don't have to artificially exclude the case that $\mu = 0$.

The essential point is the following: you want to compute the set of points satisfying $x^T C_1 x = x^T C_2 x = 0$. This is the same as computing the set of points satisfying

$$x^T (a C_1 + b C_2) x = x^T (c C_1 + d C_2) x$$

for any $a, b, c, d$ such that $ad - bc \neq 0$. Consequently it is potentially possible to simplify the problem a lot by choosing $a, b, c, d$ carefully and this determinant calculation is just one way to do that. (Also it generalizes to conics in higher dimensions.)

Qiaochu Yuan
  • 419,620