# 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?

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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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@EMACK: Another nice similar problem can be found here, mathforum.org/library/drmath/view/54510.html. – Babak S. Nov 28 '12 at 12:46
There is an easier way of finding a, without using baskara. $$a^2-a+b^2-b=(a+b)(a-b)-(a-b)=0$$ then since the solution $a=b$ is trivial we can consider $a$ diferent from $b$ and take the $a-b$ out of the equation. $$a+b-1=0$$ $$a=1-b$$ – Ivan Lerner Nov 28 '12 at 13:56
Nice way +1.... – DonAntonio Nov 28 '12 at 14:07
@IvanLerner How do we $a=b$, don't they have to be two distinct t-values? And if $a=b$ is true, then wouldn't be have $(a+b)\cdot 0-0=0$? Why can you just "take out" the $(a-b)$? – Mack Dec 3 '12 at 17:15
That is exactly why I assumed $a$ is different than $b$. If they were equal we could not take it out, because what I meant by taking out was $$(a+b)(a-b)-(a-b)=(a-b)(a+b-1)=0$$ Dividing everything by $(a-b)$ gives us $a+b-1=\frac{0}{(a-b)}$ The right side is only possible if $(a-b)$ is different than zero. Thats why, since we are looking for $a$ different than $b$, we can "take it out". If $a=b$ thenyour equation is correct and we don't get any new information, so I just ruled that out. – Ivan Lerner Dec 3 '12 at 20:28

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

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