## Parabola Geometry

### Equations

Some convenient equations for parabolas are:

x² = 4fy

where f = the focal length of the parabola. Click here to see where we get this equation from.

y = x²/4f = x²/2R

Click here to see why R = 2f. ### Proof that a Parabola focusses parallel rays of light to a Point

How do we know that a parabola is the right shape? Light from a distant star arrives in parallel rays. We want the mirror to focus the light to a point, just like a lens. So here is the geometric proof:

The general equation for a parabola is quadratic: y = a x²

Calculus tells us that the slope of such a curve is given by: y' = 2ax

But we know that the slope is just the rise over the run, which is the same as the tangent. So y'= tanq, thus we have tanq = 2ax Now tan 2q = x/d
so the distance d = x/ tan 2q

and, by trigonometric relationship, tan 2q = 2 tan q/(1-tan²q)
Subsitituting from above tanq = 2ax, we have
tan 2q = 2 (2ax) / [1-(2ax)²] = 4ax/(1-4a²x²)

So d = x (1-4a²x²) / 4ax = (1/4a) - ax²

Thus f = d+ax² = 1/4a and is invariant with x. Q.E.D.

It also gives a convenient formula for a parabola: y = x² /4f

## Radius of Curvature

How do we know that the radius of curvature R is twice the focal length?

Calculus tells us that for any curve, the radius of curvature of the curve at a particular point is given by
R(x) = [(1+y'(x)²)3/2]/y"(x)

where y' = the first derivative at the point
and y" = the second derivative at the point
click here for a derivation of this if it isn't in your calculus textbook.

For the parabola y(x) = x²/2f, the first derivative is:
y'(x) = x/2f = x/R

and the 2nd derivative is y"(x) = 1/2f = 1/R

Plugging this into the radius of curvature expression gives:
R(x) = [1+(x/2f)²]3/2/(1/2f) = 2f [1+x²/4f²]3/2

At the very center of the parabola, where x=0, we have
R(0) = 2f = R

Now, most telescope mirrors are shallow, so they have long focal lengths, and thus long radii of curvature. If you plug in the actual numbers, the radius of curvature of the outer edge is only slightly longer than the radius of curvature of the center. But it is precisely this difference that the Foucault test attempts to measure.