Question

设椭圆x^2/a^2+y^2/b^2=1(a>b>0)的两个焦点F1,F2,长轴两端点为A1A2

Answer 1

解 由余弦定理 PF1^2 +PF2^2 -2PF1*PF2*cos60 = F1F2^2

→ PF1*PF2 = PF1^2+PF2^2-F1F2^2 = (PF1+PF2)^2 - F1F2^2-2PF1PF2

→ PF1PF= (4a^2-4c^2)/3

→ △F1PF2的面积 = PF1PFsin60/2 = √3 * (a^2-c^2)

QF1^2 +QF2^2 = F1F2^2

→ (QF1+QF2)^2 - 2QF1QF2 = F1F2^2

→ 4(a^2-c^2) = 2QF1QF2 ≤ (QF1+QF2)^2/2=2a^2

→ 2a^2 ≤ 4c^2

→ e^2 ≥ 1/2 → e∈[√2/2,1)