<s>[INST] Coordinates of the center of curvature of $y^2 = 4px$ 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 $(3x+2p,\frac{-y^3}{4p^2})$.
 [/INST] 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 $\frac{1}{x'}$.
</s>