Hosted by The Math Forum

# Locus of the Focus

MacPOW Home || Math Forum POWs || Search MacPOW

## Solutions

The locus of the focus of a parabola as it rolls is a catenary.

There are several ways to do this. Here is a solution by Piotr Zielinski. Mine follows, using Mathematica for some integration.

The answer is 4y = cosh 4x.

The only proof I can come up with is calculus-based; here's a sketch.

For simplicity, consider a parabola that is four times bigger, with the focus initially at (0,1), so that we end up with y = cosh x.

Let F be the focus of the parabola, and A be the "point at infinity" along its axis of symmetry.

Consider the standard coordinate system OXY, and let C the point of contact of the parabola with OX. (Points X,Y are "at infinity.")

From the properties of the focus, we get the following angles:

∠ACX = ∠FCO = z, say, and ∠AFC = 2z

Consider the distance between the focus and the point of contact:

|FC| = 2/(1 - cos ∠AFC) = 1/sin2 z

Let (x,y) be the coordinates of F in the OXY coordinate system:

 y = |FC| sin ∠FCO = |FC| sin z = 1/sin z (1) As the parabola rolls along OX, the focus moves so that the vector dF is always perpendicular to FC, so dx/dy = sin z/cos z (2)

Combining (1) and (2), we get

dx = dy/√(y2 - 1),

whose only solution containing (x,y) = (0,1) is

y = cosh x.

And my method:

© Copyright 2008 Stan Wagon. Reproduced with permission.

[Privacy Policy] [Terms of Use]

© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/

22 September 2008