= Solution
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> is
$$
\boxed{x=\frac{2t}{1-t^2},\qquad y=\frac{1+t^2}{1-t^2}.}
$$
The 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 $t=\pm1$ 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 curve>
$$
\boxed{x=t^2,\qquad y=t^3}
$$
has inverse $t=y/x$ where $x\ne0$, and $t=0$ gives the <cusp>. Indeed, if $x\ne0$ and $y^2=x^3$, then $(y/x)^2=x$ and $(y/x)^3=y$. \b[Both curves admit rational parametrizations], although the second has a singular point.
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