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

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.