Reflections in Parabolic Mirrors
Date: 05/14/2005 at 22:16:43 From: Jessica Subject: reflective property of the parabola I am currently working on a math research paper, and I need to explain the reflective property of the parabola, such as in automobile headlights. I've done searches and have found some explanations, but I don't fully understand why it works. Can you please explain it to me in the simplest way? Thank you so much.
Date: 05/15/2005 at 10:38:35 From: Doctor Jerry Subject: Re: reflective property of the parabola Hello Jessica, I'll be referring to the image at http://mathforum.org/dr.math/gifs/parref.jpg The parabola has the equation y^2 = 4px, and I've chosen an arbitrary point (x0,y0) at which to prove the reflective property, that is, to show that the angles a and b are equal. On the diagram two angles are marked as b. The lower one is the one involved in the reflective property. The upper one--a vertical angle of the other--is useful in the proof. The focus of the parabola is the point (p,0). We need one fact: the tangent line to the parabola at (x0,y0) has slope 2p/y0. This can be easily worked out with calculus. Or one can find this result in Apolonius' book on conics. Using the relationship between the slope of a line and the angle of inclination of that line, we see that tan(a) = 2p/y0. The slope of the line from (p,0) to (x0,y0) is y0/(x0-p). So, tan(a+b) = y0/(x0-p). Because b = (a+b) - a, we can use a trig formula for the tangent of a difference and write tan(b) = tan[(a+b) - a] = [tan(a+b) - tan(a)]/[1+tan(a+b)*tan(a)] In this expression, we can replace tan(a+b) by its equal of y0/(x0-p) and tan(a) by its equal of 2p/y0. After these substitutions we can replace x0 by y0^2/(4p) and work with this messy fraction a bit. We find that it reduces to 2p/y0. So, recalling that tan(a) = 2p/y0, we have shown that angles a and b are equal. A beam of light directed from (p,0) toward (x0,y0) will be reflected from (x0,y0) along the horizontal line through (x0,y0). So, if a bulb is at the focus, the rays reflected from the parabolic reflector will be parallel to the main axis of the flashlight. Or, if parallel starlight comes into a parabolic reflector, it is all collected at the focus. Please write back if my comments are not clear or if you still have questions. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/
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