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So, I have to compute the evolute of the curve:

$y^{2} = 2px$

But I have to do that by computing the surronding of the normal rect family. So I start taking the positive side of the function and finding the normal rects:

$f(x) = \sqrt{2px}$

$f'(x) = \dfrac{\sqrt{p}}{\sqrt{2x}}$

Normal rects: $y -f(x_{0}) = \dfrac{-1}{f'(x_{0})}(x-x_{0})$

Now replacing $f(x_{0})$ and $f'(x_{0})$:

$y - \sqrt{2px_{0}} = \dfrac{\sqrt{2x_{0}}}{\sqrt{p}}(x_{0} - x)$

Now, acording to my book, if I have a family of curves $\phi(y,x,x_{0}) = 0$ I can compute the surronding of that family of curves by solving:

$\phi(y,x,x_{0}) = 0$ and $\dfrac{\partial\phi(y,x,x_{0})}{\partial x_{0}} = 0$

So, for my example I'll have to solve:

$\phi(y,x,x_{0}) = y + \dfrac{\sqrt{2x_{0}}}{\sqrt{p}}(x_{0} - x) - \sqrt{2px_{0}} = 0$

$\dfrac{\partial\phi(y,x,x_{0})}{\partial x_{0}} = \sqrt{\dfrac{2x_{0}}{p}} + \sqrt{\dfrac{p}{2x_{0}}} + \dfrac{(x_{0}-x)}{\sqrt{2px_{0}}} = 0$

Now, the problem is how to solve that system?

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Should there be a minus sign in front of $\dfrac{\sqrt{2x_{0}}}{\sqrt{p}}(x_{0} - x)$, second from last equation? –  Graham Hesketh Apr 26 '13 at 15:59
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