一道高中数学椭圆题

由题知F2(c,0),P(a^2/c,y) y是设的
|F1F2|=|F2P|
|F1F2|^2=|F2P|^2
(2c)^2=(a^2/c-c)^2+y^2
e=c/a代入得
(3e^2+2-1/e^2)a^2=y^2>=o a^2>0
3e^2+2-1/e^2>=0
3(e^2)^2+2e^2-1>=0
用一元二次不等式的解法得e^2>=1/3,或e^2<=-1(舍去)
解e^2>=1/3得e>=根号3/3,或e<=-根号3/3(舍去)

又0<e<1,所以1>e>=根号3/3

|PF2|=|F1F2| P(a^2/c,y)
4c^2=(a^2/c-c)^2+y^2
y^2=4c^2-(a^2/c-c)^2>=0
4c^2-(a^4/c^2+c^2-2a^2)>=0
3c^2+2a^2-a^4/c^2>=0
3e^2+2-1/e^2>=0
e^2>=1/3
√3/3<=e<1