I'm trying to find the coordinates of the center of curvature at any point of
$$ y^2 = 4px $$
I start with the first and second derivatives
$$ y' = \sqrt{\frac{p}{x}} \\ y'' = - \frac{1}{2} \sqrt{\frac{p}{x^3}} \\ $$
I then use the curvature formula
$$ K = \frac{|y''|}{(1 + y'^2)^{3/2}} = \frac{\sqrt{p}}{2(x+p)^{3/2}} $$
I then get the radius of curvature from the reciprocal of the curvature
$$ \rho = \frac{2(x+p)^{3/2}}{\sqrt{p}} $$
Now, I think I need to find the direction of the radius to get any point. I know that the tangent of the curve is given by the first derivative
$$ \tan \phi = \sqrt{\frac{p}{x}} $$
I also know that the slope of the radius is perpendicular to the tangent
$$ m = -\sqrt{\frac{x}{p}} $$
That's where I'm stuck finding my way towards the answer $(3x+2p,\frac{-y^3}{4p^2})$.
Given $m = -\sqrt{\frac{x}{p}}$, let $\theta$ be the angle of your radius of curvature
$$ \cos \theta = \frac{\sqrt{p}}{\sqrt{x+p}} \\ \sin \theta = \frac{-\sqrt{x}}{\sqrt{x+p}} \\ $$
So the coordinates of the center of curvature are
$$ x = \rho \cos \theta + x = \frac{2(x+p)^{3/2}}{\sqrt{p}} \frac{\sqrt{p}}{\sqrt{x+p}} + x = 2(x+p) + x = 3x + 2p \\ y = \rho \sin \theta + y = \frac{2(x+p)^{3/2}}{\sqrt{p}} \frac{-\sqrt{x}}{\sqrt{x+p}} + y = \frac{-y(x+p)}{p} + y = \frac{-yx}{p} = \frac{-y^3}{4p^2} \\ $$
Note that the first and second derivatives are simpler with respect to $y$
$$ x' = \frac{y}{2p} \\ x'' = \frac{1}{2p} \\ $$
The curvature is calculated the same way but, be careful, the slope of the tangent is $\frac{1}{x'}$.