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

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.

For $a>0$, both coordinates are nonnegative exactly on the loop branch $t\ge 0$. It leaves the origin at $t=0$ and returns there in the limit $t\to +\infty$. Thus the loop is traced once bywith the limiting endpoint identified with the starting point. The same parameter range traces the rotated loop when $a<0$.

On the loop, $y=tx$, so $dy=tdx+xdt$ and $xdy-ydx=x^{2}dt$. The traversal is anticlockwise. The area formula from Green's theorem becomesSet $u=1+t^3$ to obtainAlthough the parameter interval is unbounded, the return to the origin closes the loop and the integral converges.

Cubing and adding the parametrized coordinates yieldsTherefore the implicit algebraic curve isFor $a\ne 0$, every nonorigin real point on this curve has $x\ne 0$ and can be recovered using $t=y/x$; the origin is already included. The equation is invariant under exchanging $x$ and $y$, so its reflection line is $\boxed{y=x}$. In the parametrization that reflection corresponds to replacing $t$ by $1/t$. If $a=0$, the implicit cubic degenerates to the line $x+y=0$, whereas the given parametrization only describes the origin, so this degenerate case must be kept separate.