How to find where a set of parametric equations crosses itself

Question

The set of parametric equations:
$x(t) = t^2 − 2$
$y(t) = t^3 − 9t + 5$

Compute $(x,y)$ where the crossing occurs.

The answer key says that $(x,y) = (7,5)$, which means that $t=3$. That's really all I have, I have no idea how to solve this problem. Can someone point me in the right direction?

Answer 1

The $t$ of interest must be of the form of $\pm t_0, t_0 \ne 0$ to make $x$ takes the same value.

$$t_0^3-9t_0+5 = -t_0^3+9t_0+5$$

$$t_0^3-9t_0=0$$

$$t_0(t_0^2-9)=0$$

Hence we know $|t_0|=3$.

Answer 2

This is a special case of the self-intersection problem for cubic Bézier curves; I wrote an answer to this general problem here. Using the specific values in this problem, the relevant linear equation is $$\begin{bmatrix}0&1\\1&-9\end{bmatrix}\begin{bmatrix}\lambda^2+\lambda\mu+\mu^2\\\lambda+\mu\end{bmatrix}=\begin{bmatrix}0\\9\end{bmatrix}$$ from which we get $\lambda^2+\lambda\mu+\mu^2=9$ and $\lambda+\mu=0$. Here $\lambda,\mu$ are the parameters of self-intersection; solving the last two equations we get $\lambda,\mu=\{-3,3\}$ and the point of self-intersection as $(7,5)$.