Let $p(x,y)\in \mathbb{R}[x,y]$ be an irreducible polynomial and $deg(p)=2$. Then $p$ defines a conic $Q$ in $\mathbb{P}^2(\mathbb{R})$ as $$Q=\{[x,y,t]\in \mathbb{P}^2(\mathbb{R})|\hat{p}(x,y,t)=0\}$$
where $\hat{p}$ is the homogenized polynomial generated from $p$.
What I want to prove is that there is a projective line in $\mathbb{P}^2(\mathbb{R})$ not intersecting $Q$
I can see that the only conics of $\mathbb{P}^2(\mathbb{R})$ which intersect all lines are:
- two intersecting lines (i.e. $x^2-y^2=0$)
- a double line ($x^2=0$)
Which clearly must both be product of two linear factors.
Is there a more elegant proof that can be generalized to a higher dimention case?
thank you