请教高二数学题（解析几何）

设椭圆方程为x^2/a^2+y^2/b^2=1
F1(c,0),F2(-c,0)(对于焦点在y轴的情况类似考虑即可)
设P(acosA,bsinA),则向量PF1=(c-acosA,-bsinA),向量PF2=(-c-acosA,-bsinA)
向量PF1·向量PF2=a^2cos^2A-c^2+b^2sin^2A
注意b^2=a^2-c^2,所以向量PF1·向量PF2=a^2cos^2A-c^2+(a^2-c^2)sin^2A
=a^2-c^2(1+sin^2A)取值范围为[c^2,3*c^2]
所以e^2<=1-e^2(1+sin^2A)<=3e^2
解出1/(4+sin^2A)<=e^2<=1/(2+sin^2A),注意0<=sin^2A<=1
所以1/5<=1/(4+sin^2A)<=e^2<=1/(2+sin^2A)<=1/2
所以根号5/5<=e<=根号2/2