Kevin Wampler wrote:

> A beam from the focus going directly horizontal should be reflected directly
> upward, which means that the parabola at that point has a slope of 1 or -1.
> Since the beam is horizontal, it gives us the y-coordinate of the focus.
> So...
>
> d/dx (a*x^2+b*x+c) = 2*a*x+b
> 2*a*x+b = 1    ==>   x = (1-b)/(2*a)
> y = a*x^2+b*x+c   ==>   y = (4*a*c-b^2+1)/(4*a)
>
> Similarly with the x-coordinate:
>
> 2*a*x+b = 0    ==>   x = -b/(2*a)
>
> So, unless I made a mistake, the focus should be at <-b/(2*a),
> (4*a*c-b^2+1)/(4*a)>
>
> David Fontaine wrote:
>
> > You want a parabolic mirror with a light at its focus, like in a
> > flashlight. I don't know how to find the focus of a parabola though, Warp
> > or somebody?

But if you use a parabola and a point light to make a parallel light, only the
light that bounces off the parabola is parallel, you have to block the rest of
the light somehow.

Dan Johnson

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