How Can R=2f Be Proven in Optics?

[NiuEr]

Help

Hi... Do you know how to prove R=2f where R is the radius of curvature and f is the focal length?

 

[HallsofIvy]

The parabola $y= \frac{1}{4f}x^2$ has focus at (0, f) so focal length f. The curvature of a function y= y(x) is given by
$$\frac {|y"|} {(1+(y')^2)^{ \frac {3}{2}}}$$

In this case, y'= (1/(2f)x and y"= 1/(2f). At x= 0, the curvature is
1/(2f) and so the radius of curvature is 2f.

 

[NiuEr]

Sorry, I can't see the image of the graph. Can you please send it again? thank you very much!