# Parabolic problem

• Feb 17th 2007, 02:04 PM
gretchen
Parabolic problem
If I had a parabolic reflector that had a width w=20 feet and a depth d= 4 feet, how far from the center would the focal point be?
• Feb 17th 2007, 04:08 PM
ticbol
Quote:

Originally Posted by gretchen
If I had a parabolic reflector that had a width w=20 feet and a depth d= 4 feet, how far from the center would the focal point be?

The focal point, or focus, is "p" away from the vertex.

Let us position the parabolic reflector such that its axis of symmetry is horizontal, and its vertex is at (0,0).
The standard form of the equation of the parabola of the reflector is
(x-0)^2 = [1/(4p)](y-0)^2
x = [1/(4p)](y^2) --------(i)

The upper corner of the "width" of the reflector is at (4,10).
The lower corner is at (4,-10)

At point (4,10),
4 = [1/(4p)](10^2)
4 = 25/p
p = 25/4
p = 6.25 ft = 6ft and 3in.

Now, "...how far from the center would the focal point be?"

If by "center" you mean the center of the "width", then the focal point is
6.25 -4 = 2.25ft = 27 inches outside of the plane of the width, along the focal axis.

If by "center" you mean the center of the reflector's parabolic disc, or the vertex, then the focal point is 6.25ft = 6ft, 3in. = 75 inches away along the focal axis.

If by "center" you mean another point from those two above, be specific. We can locate the solved focal point relative to any point.
• Feb 18th 2007, 01:19 AM
gretchen
The center was the center of the reflector's parabolic disc, or the vertex, so the focal point is 6.25ft.

Thanks.