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!