= Solution
Work over the usual algebraically closed ground <field> of characteristic zero. Away from $x=0$, a line of slope $t=y/x$ through the singular point meets the <curve> with $t^2=x-1$. Thus
$$
\boxed{\phi(x,y)=y/x,\qquad \psi(t)=(t^2+1,\ t(t^2+1)).}
$$
Substitution verifies the defining equation. On $x\ne0$ and $t^2+1\ne0$, the two maps are inverse <morphisms>, so this is a <birational map> to $\mathbb A^1$. The two parameters $t=\pm i$ both map to the node $(0,0)$, explaining why the globally defined parametrization is not a global <isomorphism>.
Back to article page
