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### 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/
```
Associated Topics:
High School Conic Sections/Circles
High School Trigonometry

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