Question

已知椭圆x2/a2+y2/b2=1(a＞b＞0)的右焦点为F1(2,0),离心率为e.设A,B为椭圆上关于原点对称的两点

Answer 1

(2)

令A(p, q), B(-p, -q)
M((p + 2)/2, q/2), N((-p + 2)/2, -q/2)
O在以MN为直径的圆上，则OM⊥ON
OM的斜率u = (q/2)/[(p + 2)/2] = q/(p + 2)
ON的斜率v = (-q/2)/[(-p + 2)/2] = q/(p - 2)
uv = -1 = q²/(p² - 4)
p² + q² = 4
即A在圆x² + y² = 4上
c = 2, c² = 4

现在考虑圆x² + y² = 4和圆x²/a² + y²/b² = 1椭圆的交点
b²x² + a²y² - a²b² = 0
(a² - c²)x² + a²(4 - x²) - a²(a² - c²) = 0
-c²x² + a²(4 - a² + c²) = 0
c²x² = a²(4 - a² + c²)
4x² = a²(8 - a²)
x² = a²(8 - a²)/4
不妨在第一象限交点P, OP的斜率k ≥ √3, k = √3时, 倾斜角为π/3, P的横坐标为2cos(π/3) = 1, 所以0 < x² = a²(8 - a²)/4 ≤ 1
a²(8 - a²) > 0 => 0 < a² < 8 (i)
a²(8 - a²)/4 ≤ 1 => 0 < a² ≤ 4 - 2√3 (此时a < b, 舍去)
或4 + 2√3 ≤ a² (ii)
结合(i)(ii): 4 + 2√3 ≤ a² < 8
1/8 < 1/a² ≤ 1/(4 + 2√3)
4/8 < 4/a² ≤ 4/(4 + 2√3)
1/2 < e² ≤ 4 - 2√3 = (√3 - 1)²
√2/2 < e ≤ √3 - 1