<s>[INST] Finding Where A Parametric Curve Intersects Itself 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: $a^2-a=b^2-b$ and $a^3-3a-1=b^3-3b-1$.
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?
 [/INST] 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 $\,t\,$!
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