[Setiqwest]

After messing with some figures with paraboliods (my interest being in telescopes) I ran into a problem with the shape. Setting the origin to 0,0, the formula is ax^2. My interest was where the light would come to a point, or the focal length. I brought out my graph calc. And calculated near tangents by selecting two points that were one thousandths of an inch to create the tangent. I know this isn't exact, but pretty damn close. Then I found the perpendicular and found that line and graphed it on a piece of paper. I selected two points on the parabola so I could find where two perpendicula lines came across Y-axis. Then I bisected the two perpendicular lines to the tangent too which I thought would all come to one point: Thinking the angle of incident equalls angle of reflection. But too little avail, the points didn't come together. I was using the coeffecient of .03 for A. I'm trying to obtain a specific foucault ratio or f/ of 7. I do know that .03 directly effects the focal point but I don't know if it is directly proportional. Without actually having the mirror, how would I find the focal point on graph paper? And why was it my focal points weren't coming to one point?

[CommDweeb]

Parabolas perfectly focus light sources that are infinately far away. The closer the light source, the more the distortion. This is not a problem when you are star gazing. A paraboloid is ideal for astronomy. The easiest way is to find the focal point of a parabaloid is to start with a light source that is infinately far away. This means that all incoming light rays are parallel to the why axis. So, for any given point P on a parabola, do the following.

A) Construct a line L1 parallel to the Y axis through P
B) Construct a line L2 perpendicular to the parabola at point P
C) Construct a line L3 that mirrors exactly the angle betweenL1 and L2 using the L2 (the perpendicular) as the mirror axis.

If everything is done right, L3 will intercept the why axis at the focal point.

There is an easier way. The slope of a parabola a*x^2 at any point is simply 2*a*x. When this slope equals exactly 1, the reflection will be perpendicular to the incident ray, or parallel to the x axis. This reflection will pass through the focal point, so if we can find this point, its why value will equal the focal length. Start with 2*a*x=1. Solving for x we get x=1/(2*a). Substituting in the original equation, we get y=a*(1/(2*a))^2. Squaring this reduces to y=a*(1/(4*a^2)) which reduces to y=1/(4*a) at the focal point. This makes the focal length (fl) fl=1/(4*a).

I hope this helps. :)