Finding the Value of C for Orthogonal Intersection of Two Curves

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Homework Statement


Two curves intersect orthogonally when their tangent lines at each point of intersection are perpendicular. Suppose C is a positive number. The curves y=Cx^2 and y=(1/x^2) intersect twice. Find C so that the curves intersect orthogonally. For that value of C, sketch both curves when -2 < x < 2 and 0 < x < 4


Homework Equations


y=(1/x^2)
y'=(-2/x^3)


The Attempt at a Solution



The tangent line at the the other equation (y=Cx^2) must be the opposite reciprocal of the tangent line y'=(-2/x^3), therefore it must =(x^3/2) this has to equal the derivative of the equation y=(Cx^2) for this equation to have this derivative C must = (x^2/8) because when subsituted into the equation y=(Cx^2)you get y=(x^2/8)(x^2) which is equal to y=(x^4/8) when you find its derivative you get y'= (x^3/2).
I'm not sure if I am going about doing this correctly if anyone can point me in the right direction please give me anything you have to offer. thanks
 
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At which point(s) do they intersect?
 
(1.4,.5), (-1.4,.5)
 
anthonym44 said:
(1.4,.5), (-1.4,.5)
How is it that these points do not involve terms with C? (I don't understand how or why they are independent of C.)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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