Elliptic left directrix x=-a? /c
The vertical lines passing through A and B respectively are left alignment lines, and the vertical feet are A' and B' respectively.
Then b is BA 1, perpendicular to AA' through AA' and A 1.
Then Δ ba1a is an isosceles right triangle (because the inclination of AB is 45).
Let | BF | = m(m >;; 0) Then |AF|=3m
Available |AA'|=3m/e, |BB'|=m/e
(Ellipse property: the ratio of the distance from any point to the focus to the distance from the same side directrix is equal to E)
|AA 1|=|AA'|-|BB'|=2m/e
δδba 1A:| aa 1 | = 2m/e,|AB|=4m。
|AB|=(√2)|AA 1|
Get 4m=(√2)(2m/e).
The solution is e=√2/2.
So choose C.
I hope I can help you!