Parametric equations and tangent(s)

[underivy]

Homework Statement

At what point does the curve $$x = (1-2cos^2(t)$$, $$y=(tan(t))(1-2cos^2(t))$$ cross itself? Find equations of both tangents at that point.

Homework Equations

The Attempt at a Solution

To begin, I figured that x=y when it crosses itself, so I set x=y and got $$1 = tan(t)$$, so $$t=\pi/4$$

Next, I decided to solve for dy/dx:

$$dx/dt=4cos(t)sin(t)$$
$$dy/dt=4sin^2(t)+sec^2(t)-2$$

$$dy/dx=\frac{(dy/dt)}{(dx/dt)}$$

At that point, I plugged $$\pi/4$$ into my eqn for dy/dx and got that the slope at that point is $$1$$

Additionally, I solved for x and y at $$\pi/4$$ and got that both x and y are 0 when $$t=\pi/4$$. Using these points and the slope, I got that the equation of the tangent line is:

$$y=x$$BUT

That's just one of them. If the curve is crossing itself once, it clearly has two tangent lines at that point, but I'm not sure how to find it. Is the slope of that line just $$-1$$? I guess it might be, but it doesn't seem right to just assume that it is.

Should I have said that $$t=n\pi + \pi/4$$? If that is the case, would I just say that $$t=5\pi/4$$ AND $$t=\pi/4$$? Of course, that assumes that I did the rest of the problem properly, and I'm not even sure if I did. I was very ill all last week, so I have no lecture/recitation notes to go by, and my book doesn't have very many examples for me to practice from.

I'm sorry if this is a bad question. :( Thanks for looking!

 

[roadrunner]

im not 100% on this, but i don't think its when x=y
i think it means there are two different values of t that give the same x and y values?
but i could be wrong

 

[EnumaElish]

Given x = f(t), y = g(t), I think the question is to find t0 and t1 such that f(t0) = f(t1) and g(t0) = g(t1). Or to find t and s so that f(t) = f(t+s) and g(t) = g(t+s).

 

[HallsofIvy]

Exactly. Not "y= x" but f(t)= f(s) , g(t)= g(s). (Yes, you could find s such that the second point is t+ s, but I think that's unnecessary and too complicated. And I expect EnumElish eant f(t)= f(s, g(t)= g(s) anyway.)

Shouldn't be too difficult to find all solutions to f(t)= 1- 2cos²(t)= 1- 2cos²(s)= f(s). Now which of those also satisfy g(t)= tan(t)(1-2cos²(t))= tan(s)(1- 2cos²(s))?

 

[roadrunner]

delete this post please

 

[underivy]

Ah, I think I see now. Thanks for the help!

I guess it wasn't enough to simply say:

$$(1-2cos^2(t) = (tan(t))(1-2cos^2(t))$$ and solve for t like I was doing. However, the answer I got for that came out close to what I got for $$1- 2cos2(t)= 1- 2cos2(s)$$

I appreciate the help. :)