Proof of Reflective Property of the HyperbolaDate: 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. Steps: 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. Conclusion: 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 http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/