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Angles of Reflection

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Date: 03/13/2001 at 01:21:20
From: Nathan Niederkohr
Subject: Proving why one ray from focus in an ellipse reflects to the
other focus

I am using a tangent line on an ellipse to get angles of reflection.
I am trying to prove that the angle at the tangent line of a ray from
one focus is equal to the angle at the tangent line of the ray being
reflected at the other focus. Basically I am trying to prove the
incoming angle of a ray from one focus within a ellipse is equal to
the angle of the outgoing ray to the other focus. I am tring to prove
the angles at the edge of the ellipse with a tangent line equal to
each other.

I would really appreciate a response, because I am having a lot of
difficulty in trying to prove this.

Thank You, Nathan
```

```
Date: 03/13/2001 at 12:37:00
From: Doctor Rob
Subject: Re: Proving why one ray from focus in an ellipse reflects to
the other focus

Thanks for writing to Ask Dr. Math, Nathan.

Many of the formulas you will need appear on the following Web page
from our Frequently Asked Questions (FAQ):

Analytic Geometry Formulas
http://mathforum.org/dr.math/faq/formulas/faq.analygeom_2.html

Let the ellipse be in standard position, so having the equation

x^2/a^2 + y^2/b^2 = 1,   a >= b > 0.

Let c be defined by

c = sqrt(a^2-b^2).

Then the foci are at F1(-c,0) and F2(c,0). Let a point P on the
ellipse have coordinates (x0,y0). Then the slope of the tangent line
at P is

m0 = -b^2*x0/(a^2*y0),

provided y0 isn't 0. (If it is, the tangent line is vertical, and
you are at the left or right vertex of the ellipse. Here the desired
property is obvious, and the angle of incidence of both focal lines
is a right angle.)

The slopes of F1P and F2P are, respectively,

m1 = y0/(x0+c),
m2 = y0/(x0-c),

provided neither line is vertical. (We'll deal with that case later.)
Now the angle between F1P and the tangent line is

theta1 = arctan([m1-m0]/[1+m1*m0]),

and the angle between F2P and the tangent line is

theta2 = arctan([m0-m2]/[1+m0*m2]).

Substitute the expressions for m0, m1, and m2 into these equations,
and you'll find, after some manipulation, that both these angles are
equal to

theta1 = theta2 = arctan(b^2/[c*y0]).

If PF2 is vertical, then x0 = c, y0 = +-b^2/a, m1 = +-b^2/(2*a*c),
m2 is infinite, m0 = -+c/a, and you get

theta1 = theta2 = arctan(+-c/a).

If PF1 is vertical, then x0 = -c, y0 = +-b^2/1, m1 = -+b^2/(2*a*c),
m1 is infinite, m2 = +-c/a, and you get

theta1 = theta2 = arctan(-+c/a).

Thus in any case theta1 = theta2.

I leave the computational details to you.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Euclidean Geometry

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