Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 10 / c

For $ab\ne 0$, elimination of the trigonometric parameter givesAs real $\theta$ varies, the whole ellipse is traced, with semiaxes $|a|,|b|$. Signs of $a,b$ affect the traversal direction but not the set. If $a=0$, the locus is the vertical segment $x=0$, $|y|\le |b|$; if $b=0$, it is the horizontal segment $y=0$, $|x|\le |a|$. Both zero give only the origin.

For $ab\ne 0$, the hyperbolic identity gives $x^2/a^2-y^2/b^2=1$. The hyperbolic parametrization of a hyperbola also imposes $x/a=\cosh \theta \ge 1$, so the exact traced locus isIt is only one branch: the right branch for $a>0$ and the left branch for $a<0$. Its vertex is $(a,0)$ and its asymptotes are $y=\pm (b/a)x$. If $a=0$, $b\ne 0$, the entire vertical axis is traced. If $b=0$, $a\ne 0$, the locus is the ray $y=0$, $x/a\ge 1$. Both zero give the origin.