Question

设F1 F2分别是椭圆x^2/a^2+y^2/b^2=1(a>b>0)的左右焦点,若在其右准线上存在点P,使PF1的中垂线过点F2，求

求椭圆离心率的取值范围。答案是【根号3/3,1) 要详细过程

Answer 1

右准线: x = a²/c
F₁(-c, 0), F₂(c, 0)
令P(a²/c, p)
PF₁的斜率为k = p/(a²/c + c)，
PF1的中垂线斜率为k' = -1/k = (a²/c + c)/p = (a² + c²)/(pc)
PF1的中点M((a²/c - c)/2, p/2)
PF1的中垂线: y - p/2 = [(a² + c²)/(pc)][x - (a²/c - c)/2]
中垂线过点F₂(c, 0): 0 - p/2 = [(a² + c²)/(pc)][c - (a²/c - c)/2]
整理得: p²c² = (a² + c²)(3c² - a²)
p²/c² = (a²/c² + 1)(3 - a²/c²) = (1/e² + 1)(3 - 1/e²) ≥ 0
(1 + e²)(3e² - 1) ≥ 0
1 + e² > 0
3e² - 1 ≥ 0
e² ≥ 1/3
e ≥ 1/√3
又椭圆的离心率e < 1
1/√3 ≤ e < 1