Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 22 / 1 / i

For the first algebraic curve, a hyperbola, use the line $y=1+tx$ through $(0,1)$. Substitution and cancellation of the known intersection give $x((1-t^2)x-2t)=0$. Thus a rational parametrization of an algebraic curve isThe identity $y^2-x^2=1$ follows immediately. Away from $(0,1)$ its inverse is $t=(y-1)/x$; the exceptional point is recovered at $t=0$. The other point with $x=0$, namely $(0,-1)$, corresponds to $t=\infty$, while $t=\pm 1$ gives the points at infinity on the projective closure. This explains the exceptional parameters rather than discarding them.

For the second algebraic curve, the rational parametrization of an algebraic curvehas inverse $t=y/x$ where $x\ne 0$, and $t=0$ gives the cusp. Indeed, if $x\ne 0$ and $y^2=x^3$, then $(y/x{)}^2=x$ and $(y/x{)}^3=y$. Both curves admit rational parametrizations, although the second has a singular point.