intersection of real quadrics in P2

Question

Given two real quadrics in $\mathbb P^2$ without common components, then we know that they have at most 4 intersection points.

Suppose that the quadrics are not degenerate (I don't know if this is necessary) and that they are in the form

$$P_1(x,y) = x + Q_1(x,y) \qquad P_2(x,y) = y + Q_2(x,y)$$

where $Q_1$ and $Q_2$ are homogeneous of degree 2, and $Q_1(x,y)=Q_2(x,y)=0$ implies $x=y=0$.

Obviously, they have the common solution $(0,0)$ with multiplicity 1, and they do not intersect on the infinite line thanks to the hypothesis on $Q_1,Q_2$.

I know that there is at least an other intersection between the two quadrics, but I don't know how to prove it. Geometrically, I can see it, because the only quadrics are parabolas, circles and hyperboles, and if two of them have a simple intersection, then they have an other one. How can I prove it without doing all the computations case by case?

Answer 1

What if you have $f(x)=-3x^2$ and $g(x)=\frac{1}{2}x^2$? Then you have a single intersection at $(0,0)$ but no intersections anywhere else.

Answer 2

You can show that the pencil of conics must contain a degenerate line pair. This follows from the generalized eigenvalue problem involving your two 3x3 conic matrix representations.

$$ det(P_1 + λP_2)=0 $$

This is a cubic equation for your pencil of quadrics. This is guaranteed to have at least one real solution for λ. Such a solution is a degenerate conic, a pair of lines, and one of the lines in the pair must intersect at the origin and then must also intersect the quadric again on its way "out". Such a line is guaranteed to have two real intersection points if you know it has one real intersection point.

All full rank quadrics in P2 are projectively equivalent to a circle. So if a line goes in, it must go out.