# How would I do this? I have the answer and most steps but....

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• Mar 10th 2011, 02:29 AM
twittytwitter
How would I do this? I have the answer and most steps but....
Let A be a positive number and define a curve parametrically by
x(t)=t^3-t and y(t)=A/(1+t^2).
Note that there is one value of A for which the "self-intersection" of the curve is perpendicular. Please find this A.

I know the answer is 4, but....I'm having trouble justifying the point of intersection. I know it occurs for t=1 and t=-1, and we get (0, A/2) as the point BUT how do I show this formally? Obviously if t=1 or -1, we get the same point (x,y), but that's just by intuition. How would I do it if it were not so obviously what t values to use? That is, how do I actually find the coordinates of the intersection? I hope this question makes sense?

I don't need help after that, but here would be the rest:
I know after that dy/dx=(dy/dt)/(dx/dt). Then we plug in t=1 and t=-1 to get m1 and m2.
I know I will need m1*m2=-1 since perpendicular.
So (A/4)(-A/4)=-1, and so, A=4
• Mar 10th 2011, 06:11 AM
zzzoak
Please try to solve the system

\$\displaystyle
x(t_1)=x(t_2)
\$

and

\$\displaystyle
y(t_1)=y(t_2)
\$