p=(0,0), C: y=x^2, D: y=2x^2. Using Bezout's theorem and symmetries, show that ip(C,D)=2. Show this one more time using a parametrization of the conic.
First off, Bezout's Theorem says there are 4 points of intersection so if I could use symmetry to find the other 2 I could prove that the point at (0,0) must have multiplicity 2 but I don't know how to do this.
As for the parametrization, I really don't even know where to start. What variables do I need to substitute and what am I trying to solve for and show here?
First of all, in order to use Bezout's theorem to say we have 4 points of intersection we need to be looking at a projective curve. We can do this by homogenizing the equations to get $yz = x^2$ and $yz = 2x^2$, and the point $p$ we care about is $[0,0,1]$ in projective coordinates. Now we can also easily see that $[0,1,0]$ is also a solution, which is "at infinity" in the affine picture. Moreover if both $y$ and $z$ take nonzero values then see we can't have any solutions as $x^2 \ne 2x^2$ for $x\ne 0$.
So we see that the multiplicities of these two intersection points needs to sum to 4. But now we can observe that there is a symmetry between $y$ and $z$ in our equation, in particular if we look at our second intersection in the affine chart where $y = 1$ it looks like $z = x^2$ and $z = 2x^2$ at $x = z = 0$ which obviously looks the same as the intersection point we started with. Hence they both need to have intersection multiplicity 2.