Find the point where the curve $\big(x(t),y(t)\big)=(t^2-1,t^3-t)$ crosses itself

Question

Consider the curve defined by $x(t)=t^2-1$ and $y(t)=t^3-t.$

Find the point where the curve crosses itself.

I know that the curve will cross itself if there are two distinct values, say $t_1$ and $t_2$, such that $x(t_1)=x(t_2)$ and $y(t_1)=y(t_2).$

I have used this and have come down to $$t_1^2=t_2^2\\\text{ and }\\t_1^3-t_1=t_2^3-t_2.$$

I'm not sure how to solve these for $t_1$and $t_2$ though.

Answer 1

As remarked in the comments, the first equation leaves out two possibilities: \begin{equation} t_1=\pm t_2 \end{equation} $t_1=t_2$ is trivial, so look at the other. Plugging in $t_1=-t_2$ in the second equation, you get \begin{equation} t_1^3=t_1 \end{equation} which leaves you with three values, $t=-1,0,1$. So either $t_1=t_2=0$ (trivial), or $t_1=1$, $t_2=-1$: the solution only makes sense if $t_1\neq t_2$. So plug in either $t=1$ or $t=-1$ to get the self-intersection point!