I'm having a little trouble finding the radius of curvature for problems of the following form

$\displaystyle y^n=f(x)$

For examples. I'm asked to find the radius of curvature at the point (0,0) for the curve $\displaystyle y^2=4ax$

So

$\displaystyle 2y\, y' = 4a \implies y' = 2ay^{-1}$

and

$\displaystyle y'' = -2ay^{-2}\, y'$

So

$\displaystyle \rho = \frac{[1+(y')^2]^{\frac{3}{2}}}{y''} = \frac{[1+4a^2y^{-2}]^{\frac{3}{2}}}{-2ay^{-2}\, y'}$

it's at this point where I'm sure I've messed up. If anyone could tell me how to solve these types of questions I'd be very grateful

Regards

Stonehambey