# F=R/2 approximation

1. Aug 1, 2010

### azaharak

Hi

For small angles or points near the vertex of a parabola we can approximate a parabolic surface with a circle. The focus of the parabola is a unique point specifically for optics (Parallel light will converge at the focus), and vice versa.

Has anyone come across an derivation that shows in the limit of small angles or points near its vertex, the focus of a parabola is equivalent to R/2 where R is the radius of a circle used to approximate the parabola near its vertex.

I've seen the derivation where we obtain
1/s+1/s'=2/R,

but they we always conclude that 2/R must be 1/F.

I was wondering if there is a derivation starting with the focal point of a parabola and then approximates to a circle.

2. Aug 2, 2010

### HallsofIvy

Let the parabola be given by $y= x^2/(4f)$, so that the focal length is f, and let the approximating circle be given by $x^2+ (y- r)^2= r^2$ so that the circle and parabola both pass through (0, 0).

The first derivative for the parabola is $y'= 2x/(4f)$ which is 0 at x= 0. For the circle, the first derivative is given by $2x+ 2(y- r)y'= 0$ or $-2ry'= 0$ at (0, 0) so the derivative there is 0.

The second derivative for the parabola is $y'= 1/(2f)$ for all x. For the circle, the second derivative is given by $2+ 2y'+ 2(y- r)y''= 0$ or $2- 2ry''= 0$ when y= y'= 0 so y''= 1/r at (0, 0).

In order that the second derivatives at (0, 0) be the same, so that they have the same curvature there (this is called the "osculating" circle- look up "osculating" in a dictionary!), we must have 1/r= 1/(2f) so that r= 2f.

Since all succeeding derivatives of any quadratic function are 0, that the value, first derivative, and second derivative of two quadratic functions be the same at a point is the most stringent requirement we can place on them.