Owen Lawrence (in snippedforprivacy@quag.dido.ca) said:
 I'm not checking my work, but I'll give it a shot. First, let's
 simplify a bit by locating your vertex at the origin. Your
 parabola is then y = x^2/4a. The formula for arc length can be
 derived to be

 L = int( x1, x2 ) sqrt( 1 + (dy/dx)^2) dx, where int( x1, x2 ) is
 the definite integral from x1 to x2 on your parabola.

 dy/dx = x/2a
 (dy/dx)^2 = x^2/4a^2

 so L = int( x1, x2 ) sqrt( 1 + x^2/4a^2 ) dx
 = 1/2a int( x1, x2 ) sqrt( 4a^2 + x^2 ) dx

 I looked up the indefinite integral in my CRC math tables to be

 int() sqrt( x^2 + c^2) dx = 1/2[ x sqrt( x^2 + c^2 ) + c^2 ln( x +
 sqrt( x^2 + c^2 ) ]

 In our case c = 4a^2. Since you want your arc length to be from
 the vertex, we can let x1 = 0. So our final formula becomes:

 L = 1/4a[ x sqrt( x^2 + (2a)^4 ) + (2a)^4 ln( x + sqrt( x^2 +
 (2a)^4 ) ]

 Check the algebra. I haven't had lunch yet so maybe I'm not
 thinking clearly.
Owen...
Thank you. Even without lunch you did better than I managed. :)

Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/solar.html