Parabola equation from dimensions...?

[Externet]

Hi. Too many years since left university, cannot think on how to create the equation.

An 8 feet long string, (or rod, or sheet) forming a parabola.

The shape will be somewhat like a flattened U.

I want the focus at the center between the upper ends of the U. (Or, at the lower tip of the accent on this letter Ú )

How do I come up with the equation? -It's to build a reflector with a 8'x4' sheet and be able to mechanically support the focus between its ends-

Miguel

[BigMoosie]

Let the porabola be described by:

 

[math]x^2 = 4ay[/math]

 

Where a is the distance from the vertex to the focus. We want a specific region of x-values of this curve lets say from b to -b. That means the equation satisfies (b,a)

 

[math]b^2 = 4a^2[/math]

 

[math]b = 2a[/math]

 

This means the porabola must be 4 times wider than it is tall. This may be enough to solve your problem, but if you need specific dimensions the following is relevant:

 

The length along any curve defined by f(x) is:

 

[math]L = \int_{x_1}^{x_2} \sqrt{1+[f'(x)]^2} \cdot dx[/math]

 

So:

 

[math]f(x) = \frac{x^2}{4a}[/math]

 

[math]f'(x) = \frac{x}{2a}[/math]

 

[math]8 = \int_{-b}^b \sqrt{1+\frac{x^2}{4a^2}} \cdot dx[/math]

 

[math]8a = \int_0^{2a} \sqrt{4a^2 + x^2} \cdot dx[/math]

 

let x = 2.a.tan(w)

then: dx = 2.a.sec²(w).dw

 

[math]8a = 4a^2 \int_0^\frac{\pi}{4} sec^3(w).dw[/math]

 

[math]a = \frac{2}{\int_0^\frac{\pi}{4} sec^3(w).dw}[/math]

 

I had to use some software here to evaluate this integral, I just couldn't seem to crunch it.

 

[math]a = \frac{4}{log_e(\sqrt{2}+1)+\sqrt{2}} \approx 1.742473598...[/math]

 

So your porabola would fit inside a box measuring 6.9699 x 1.7425 (in feet).