Let $f \in k[x,y]$ be an irreducible polynomial of degree $2$, $k$ is a field, and there exist $a,b\in k$ such that $f(a,b) = 0$. I'm trying to prove that $\mathsf{frac}(k[x,y]/(f))$ is isomorphic to $k(t)$, the field of rational functions with single variable.
What I know:
$f$ is irreducible and so its prime, so the ring $k[x,y]/(f)$ is an integral domain, so the fraction field is just the quotient of its elements. But I am stuck at this point. The only thing I can think of is finding some embedding from $k[x,y]/(f)$ into $k(t)$, then $\mathsf{frac}(R)$ would be contained in $k(t)$ and go from there.
Any hints would be appreciated...
The fraction field of $R=\mathbb R[x,y]/(x^2+y^2)$ is $K=\mathbb R(y)[x]/(x^2+y^2)$, that is, $K=\mathbb C(y)$, and this is not a purely transcendental extension of $\mathbb R$ (why?). It follows that $K\not\simeq\mathbb R(t)$.