China Mathematical Competition

Question

Given an ellipse equation x^2a^2 + y^2b^2 = 1 (a > b > 0) in a plane rectangular coordinate system xOy, let A_1, A_2, F_1, F_2 be its left and right end-points, left and right focuses, respectively, and P be any point on the ellipse different from A_1, A_2. Suppose there are points Q, R satisfying QA_1 PA_1, QA_2 PA_2, RF_1 PF_1, RF_2 PF_2. Find and prove the relationship between the length of segment QR and b.

Accepted Answer

Let c = a^2 - b^2. Then A_1(-a, 0), A_2(a, 0), F_1(-c, 0), F_2(c, 0). Denote P(x_0, y_0), Q(x_1, y_1), R(x_2, y_2), where x_0^2a^2 + y_0^2b^2 = 1, y_0 0. From QA_1 PA_1, QA_2 PA_2, we have A_1Q A_1P = (x_1 + a)(x_0 + a) + y_1y_0 = 0, 1 A_2Q A_2P = (x_1 - a)(x_0 - a) + y_1y_0 = 0. 2 Subtracting ① and ②, we have 2a(x_1+x_0) = 0, i.e. x_1 = -x_0. Substituting it into ①, we get -x_0^2 + a^2 + y_1y_0 = 0, or y_1 = x_0^2 - a^2y_0. Then Q(-x_0, x_0^2 - a^2y_0). From RF_1 PF_1, RF_2 PF_2, in the same way we obtain R(-x_0, x_0^2 - c^2y_0). Therefore, |QR| = | x_0^2 - a^2y_0 - x_0^2 - c^2y_0 | = b^2|y_0|. Since |y_0| (0, b], then |QR| b, where the equality holds if and only if |y_0| = b (i.e., P(0, b)).