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?