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