S is a parabola with focus F and axis L. Three distinct normals to S pass through P. Show that the sum of the angles which these make with L less the angle which PF makes with L is a multiple of π.
Solution
We start by finding a formula for tan(A+B+C+D). Applying tan(A+B) = (tan A + tan B)/(1 - tan A tan B) twice and writing a = tan A, b = tan B, c = tan C, d = tand D, we get tan(A+B+C+D) = (∑a - ∑abc)/(1 - ∑ab + abcd).
Take the parabola as y = x2. The gradient at (k, k2) is 2k, so the normal has slope -1/2k. The normal has equation 2k3 + (1 - 2y)k - x = 0. For given x, y this is a cubic in k, so has 1 or real 3 roots. Their slopes satisfy the cubic 4x h3 + 2(1 - 2y) h2 + 1 = 0 (*).
The focus is at (0, 1/4), so the line joining (x, y) has slope (y - 1/4)/x and negative is (1/4 - y)/x. The quartic with this root and the 3 roots of (*) is (h + (y - 1/4)/x )(4x h3 + 2(1 - 2y) h2 + 1) (**). So the angles the normals make with L less the angle which PF makes with L is a multiple of π iff (∑a = ∑abc), where a, b, c, d are the slopes of the four lines and hence iff the coefficients of h3 and h in (**) are the same. But the coefficient of h3 is 2(1 - 2y) + 4x (y - 1/4)/x = 1 and the coefficient of h is 1.
© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002