The recording of Feynman's Lost Lecture does not end when the bell rings; the tape is still rolling when the physicist answers some questions afterwards. Last night, what caught my attention is his comment that there's another way to think about the reflection property of the ellipse.
If you consider the tacks and string construction, as you move about the ellipse, string is being shed on one side and taken up on the other. These have to balance out, because the length of the string is fixed.
I haven't gotten anywhere yet trying to apply this idea geometrically, but analytically it works quite well.
Consider then that we start with focus points f and f', some point P on the ellipse, and the tangent thereto. From both foci, drop altitudes to the tangent line, naming those points p and p' respectively.
The length of the string, d(f,P) + d(f',P), can be represented by a slightly more complicated expression
( d^2(f,p) + d^2(p,P) ) ^1/2 + ( d^2(f',p') + d^2(p',P) ) ^1/2
From the definition of the ellipse, we know this expression is constant along that curve. The tangent, at point P, shares the same property - a small displacement of P does not change the length [to first order]. In other words, the derivative of the expression above is 0 everywhere on the ellipse, and close enough to 0 on the tangent.
The two foci (f,f') are fixed, as are p and p' (as we displace P along the tangent - they move as we displace P along the ellipse, of course), so those terms will fall out of the derivative, as they are completely independent of P
1/2 ( d^2(f,p) + d^2(p,P) ) ^-1/2 (2d(p,P)d'(p.P)) + 1/2 ( d^2(f',p') + d^2(p',P) ) ^-1/2(2d(p',P)d'(p',P)) = 0
My apologies for the notation - I hope it is clear enough that d' is a derivative.
Rearranging some terms, we get
d(p,P)d'(p,P) / d(f,P) + d(p',P)d'(p',P) / d(f',P) = 0
There's an additional simplification available. P is on the line between p and p'. Displacing P along this line has to balance out, because the distance between p and p' is constant
d(p,p') = d(p, P) + d(P,p')
d(p,p') = d(p, P) + d(p',P)
Take the derivative of both sides
0 = d'(p,P) + d'(p',P)
therefore
d'(p',P) = - d(p,P)
So our derivative becomes
d(p,P)d'(p,P) / d(f,P) - d(p',P)d'(p,P) / d(f',P) = 0
d(p,P)d'(p,P) / d(f,P) = d(p',P)d'(p,P) / d(f',P)
d(p,P) / d(f,P) = d(p',P) / d(f',P)
Therefore angles pPf and p'Pf' have the same cosine, and are therefore the same angle. The reflection property is demonstrated.
July 1, 2003 6:41 PM
| TrackBack