Finding where the parametric curve $(t^2-t, t^3 -3t -1)$ intersects itself

Question

The problem I am working on is to find the where the curve intersects itself, using the parametric equations.

These are: $x=t^2-t$ and $y=t^3-3t-1$

For the graph to intersect itself, there must be two distinct t-values, $a$ and $b$, that when plugged into the parametric equations, produce the same output. These two t-values create two ordered-pairs that are the same.

My system of equations:$$\begin{cases}a^2-a=b^2-b\\a^3-3a-1=b^3-3b-1 . \end{cases}$$

I solved for $a$, but am not sure if I did it correctly: $a(a-1)=b^2-b$ then either $a=b^2-b$ or $a-1=b^2-b$. I would then have two values that I have to test. When I plugged in $a$, I ended up with a 6-degree polynomial, did I do something wrong?

Answer 1

Well, $\,xy=\alpha\beta\,$ doesn't necessarily means $\,x=\alpha\,\,\vee\,\,x= \beta\,$ , of course. Form a quadratic in $\,a\,$ and solve:

$$a^2-a=b^2-b\Longrightarrow a^2-a+b-b^2=0\Longrightarrow \Delta=1-4(b-b^2)=(2b-1)^2\Longrightarrow$$

$$a_{1,2}=\frac{1\pm\sqrt\Delta}{2}=\begin{cases}b\\{}\\1-b\end{cases}$$

If $\,a=b\,$ we're done, otherwise take $\,a=1-b\,$ and substitute into the second equation:

$$a^3-3a-1=b^3-3b-1\Longrightarrow 1-3b+3b^2-b^3-3+3b=b^3-3b\Longrightarrow$$

$$0=2b^3-3b^2-3b+2=(b+1)(2b^2-5b+2)\Longrightarrow b=-1\,,\,2\,,\,\frac{1}{2}$$

Thus, we have the solution:

$$t=-1\,,\,t=2\Longrightarrow (x,y)=(2,1)$$

Note that the other two possibilities give us the same $\,t\,$!

Answer 2

Hint:
$\large a^2-a=b^2-b$
$\large \left(a-\frac{1}{2}\right)^2=\left(b-\frac{1}{2}\right)^2$
$\large \left|a-\frac{1}{2}\right|=\left|b-\frac{1}{2}\right|$
$\large a,b=\frac{1}{2}\pm x, \ \ x>0$