Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 13 / a / Solution

Take $a>0$ for the sketch; changing its sign rotates the picture by $\pi$. The parametric curve passes through $(0,0)$ at $t=0$ and through $(3a/2,3a/2)$ at $t=1$. As $t\to \pm \infty$ it returns towards the origin, since $x\sim 3a/t^2$ and $y\sim 3a/t$. The positive-$t$ branch makes a loop in the first quadrant, leaving the origin tangent to the positive $x$ axis and returning tangent to the positive $y$ axis.

For a more precise sketch, differentiation givesThe loop has its largest $x$ at $t=2^{-1/3}$ and largest $y$ at $t=2^{1/3}$. Their coordinates are $a(2^{2/3},2^{1/3})$ and its reflection in $y=x$.

The parameter value $t=-1$ is excluded. Cancelling the common factor in the sum givesThus the asymptote is $\boxed{x+y+a=0}$. As $t\to -1^{-}$, $x\to +\infty$ and $y\to -\infty$; as $t\to -1^{+}$, $x\to -\infty$ and $y\to +\infty$. The branch $t<-1$ approaches the origin from the fourth quadrant as $t\to -\infty$, while $-1<t<0$ connects the other infinite branch to the origin through the second quadrant.

This is the folium of Descartes. If $a=0$, the parametrization degenerates to the origin for every defined parameter value and there is no nontrivial loop.