Find a curve which intersects any quadric I want to find an irreducible polynomial $p(X,Y)\in\mathbb C[X,Y]$ such that any quadric of the form $$aX^2+bXY+cY^2+dX+eY+f,\quad a,b,c,d,e,f\in\mathbb C.$$ has a common zero with $p$ in $\mathbb C^2$. I am not sure what to look for. Is there any theorem on it?
 A: The curve $Q$ given by $y^2=x^3-x$ works, assuming you enforce the requirement that at least one of $A$, $B$, or $C$ is nonzero. To prove this, by Bezout's theorem, our two curves will intersect at six points counted with multiplicity in $\Bbb P^2$. If the two curves don't share solutions in $\Bbb A^2$, then it must be the case that our curves intersect with multiplicity 6 on the line at infinity. Let's check that can't happen.
The unique point at infinity on $Q$ is $[0:1:0]$, so either $[0:1:0]$ is on the quadratic and our curves intersect to multiplicity 6 there, or our curves intersect in $\Bbb A^2$. If $[0:1:0]$ is on the quadratic, then $C=0$ and we need to check that there are no choices of the other five coefficients so that the intersection multiplicity is 6 at $[0:1:0]$.
To compute the intersection multiplicity at $[0:1:0]$, we're looking to compute the dimension of $$ \mathcal{O}_{Q,[0:1:0]}/(Ax^2+Bx+Dxz+Ez+Fz^2).$$ Since the local ring is a DVR, this is just the valuation of $Ax^2+Bx+Dxz+Ez+Fz^2$. It's not hard to check that $x$ has valuation $1$ and $z$ has valuation $3$ in this ring, and therefore the valuation of $Ax^2+Bx+Dxz+Ez+Fz^2$ is at most two if $A$ and $B$ are not both zero. Therefore $Q$ intersects every quadric in the affine plane.
