# Place where parametric curve itself itself

1. Mar 29, 2008

### motornoob101

1. The problem statement, all variables and given/known data
Find the place where the parametric curve intersect itself

$$x = 1-2cos^{2}t$$
$$y = tant(1-2cos^{2}t)$$

2. Relevant equations

3. The attempt at a solution
So I started with the x values..

$$1-2cos^{2}t_{1} = 1-2cos^{2}t_{2}$$

By canceling the same stuff on both sides, I got
$$cos^{2}t_{1} = cos^{2}t_{2}$$

Then I tried with y
I rewrote y in a different form.

$$y_{1} = tan t (x_{1})$$
and
$$y_{2} = tan t(x_{2})$$

This is possible since y already contains an expression for x.

Since the curve intersect itself, we know x1 must equal x2 so they cancel out.

then I am left with $$tant_{1}= tant_{2}$$

but I can't solve for t

Appreciate any help. Thanks

2. Mar 29, 2008

### HallsofIvy

Why can't you? Since tangent is periodic with period $\pi$, but one-to-one within each period, tan(t1)= tan(t2) requires that $t_1= t_2+ n\pi$. Now, which of those values satifies cos2(t1)= cos2(t2)?

3. Mar 29, 2008

### motornoob101

Ah, now I know what you meant, thanks!