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 calculusbased; 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/sin^{2} 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/√(y^{2}  1),
whose only solution containing (x,y) = (0,1) is
y = cosh x.