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