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Proof of Reflective Property of the Hyperbola

Date: 07/08/2004 at 12:31:38
From: Wendy
Subject: reflective property of the hyperbola

Let a ray of light aimed at one focus (F2) of a hyperbola hit its 
right arm at point P(x,y); it is well-known that such a ray (directed
at one focus) will "reflect" off that arm to reach the other focus
(F1) if we imagine each arm to be a mirror.

The proof of this property takes the tangent to the hyperbola's right
arm at point P, and argues that this tangent bisects the angle with
sides PF1 and PF2.  Such a proof is straightforward if you use the Law
of Cosines with the distances involved, but is algebraically tedious.
Is there a simpler proof without the need for so much manipulation?

Here's my understanding of the standard proof:
Let Q(u,0) be the point where the tangent to the hyperbola's right arm
at point P(x,y) intersects the abscissa.  Then the 5 distances 
involved are: PF1, PF2, QF1, QF2, and of course tangent line PQ.  The
requirement is to show that angle F1PQ = F2PQ (recall F1(-c,0) is one 
focus and F2(+c,0) is the other focus).  Use the Law of Cosines to
equate the cosines of the two angles, then argue they must lie in the
same quadrant to equate the angles themselves.


1. Determine that u = a*a/x; solve the equation of the hyperbola for
   y in terms of x, differentiate to get y' = tangent's slope, then
   use Point-Slope Form (using y' with point P(x,y)) to get u.

2. Use the distance formula to find tangent length PQ; QF1 = c+u and
   QF2 = c-u; PF1 = abs(a + c*x/a) and PF2 = abs(a - c*x/a).

3. Call the two angles M,N; use the Law of Cosines with the 5 sides
   calculated in step 2 and solve for cos(M) and cos(N).

4. Try to equate the algebraic expressions for cos(M) and cos(N) you
   obtained in step 3.  This is the tedious part I am trying to avoid,
   though I know about programs to automate the process.

5. Having shown that cos(M) = cos(N) in step 4, show that M,N belong
   to the same quadrant to conclude that M = N.


The incident angle the ray of light makes with the tangent is also M
(being a vertical angle), so M = N means that the angle of incidence
is equal to the angle of reflection, with the reflected ray hitting
the other focus (F1).

Date: 07/08/2004 at 13:54:22
From: Doctor Jerry
Subject: Re: reflective property of the hyperbola

Hello Wendy,

Suppose L is the tangent line at (x,y) on the hyperbola.  If a ray is
coming in from the upper right,  headed towards the left focus (-c,0)
and meets the hyperbola at (x,y), then a unit vector from (x,y)
outwards from (x,y) along this ray is

  u = [1/sqrt((x + c)^2 + y^2)]*< x + c, y >, where < p, q > is a
  vector, like p*i + q*j.

A unit vector from (x,y) upwards along the tangent line is

  v = [1/sqrt(1 + x^2*b^4/(y^2*a^4))]*< 1, x*b^2/(y*a^2) >.

The dot product of these two unit vectors is cos(alpha).  If you calculate

  tan(alpha) = 1/cos^2(alpha) - 1

you will find, using the equation of the hyperbola and a^2 + b^2 =
c^2, that

  tan(alpha) = b^2/(c*y).

If beta is the angle between -v and the vector from (x,y) to (c,0),
then, in much the same way,

  tan(beta) = b^2/(c*y).

So, QED.  I've left out some details, but I did work through the 
tan(alpha) part and found what I said.

Is this shorter?

- Doctor Jerry, The Math Forum 
Associated Topics:
College Linear Algebra

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