The radius of curvature of the curve $$(x,y(x))$$ is

$R=\dfrac{\left(1+(y'(x)\right)^2)^{\frac{3}{2}}}{\vert y''(x) \vert}$

Let

$y(x)=\dfrac{1}{4f}(x-h)^2+k,$

the parabola with vertex $$(h,k)$$ and focal length $$\vert f\vert$$.

Then

$y'(x)=\dfrac{x-h}{2f}$

and

$y''(x)=\dfrac{1}{2f}.$

Then

$R = \dfrac{\left(1+\left(\dfrac{x-h}{2f}\right)^2\right)^{\frac{3}{2}}}{\left\vert \dfrac{1}{2f} \right\vert}.$

At the vertex $$(h,k)$$ the radius of curvature is

$R = \vert 2f \vert$

and

$$\vert f \vert=\dfrac{R}{2}$$ is the focal length.

Focal length and center of curvature of parabola at vertex