Parabola: Proving Parallel Rays

Date: 05/21/2003 at 20:15:29
From: Dad
Subject: Parabola: Proving Parallel Rays

Given only the geometric definition of a parabola (all points equally 
distant from focus and directrix), provide a simple geometric proof 
that can be explained to sixth graders that all rays emanating from 
the focus reflect from the parabola parallel to the axis of symmetry.

I'm having difficulty proving to a sixth grader that the tangent line 
is an angle bisector of the isosceles triangle formed by the focus, 
the point on the parabola, and the intersection of the perpendicular 
with the directrix. This proof is necessary to continue the proof that 
the angle of incidence does indeed equal the angle of reflection.

Date: 05/22/2003 at 23:25:40
From: Doctor Peterson
Subject: Re: Parabola: Proving Parallel Rays

Hi, 

I'm not sure how much sixth graders can follow, but I think they will 
be able to see what's going on well enough to satisfy them. Let's 
take it in two steps. First, see this diagram:



Here I have constructed (in green) part of a parabola with the given 
focus and directrix; for point P I have shown the segments PF to the 
focus and PQ to the directrix, both with the same length a. In the 
second step I will be showing that line PM, joining P and the midpoint 
M of FQ, is tangent to the parabola; for now, it's enough that it 
looks right. We have an incoming light ray parallel to the axis and 
perpendicular to the directrix, so it is collinear with PQ. It strikes 
the parabola at 'angle 1', which is congruent to angle QPM. But angle 
QPM is congruent to angle FPM, 'angle 2', since PM is the median of 
isosceles triangle FPQ and consequently also the angle bisector. 
Therefore 'angle 2' is the direction in which the light will exit, 
passing through the focus.

I believe you have done this much yourself.

Now let's prove that PM is in fact the tangent:



Here I again have F, P, and Q, and have drawn PM, which I claim is the 
tangent. I define the tangent as a line that meets the parabola at P, 
and nowhere else. That is, I claim that any other point R on the 
parabola (which is at distance b from the focus and the directrix) is 
on the same side of PM as the focus.

Look at segment QR, whose length c must be greater than b because it 
is the hypotenuse of triangle QRS. Now look at triangle QRF. Since 
c>b, R must be closer to F than to Q, and consequently is on the same 
side of perpendicular bisector PM as F.

If you put R farther out on the parabola, the same will still be 
true. Thus all points on the parabola, other than P itself, are on 
the same side of PM, and the latter is the tangent we are looking for.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/