Because the tangent circle o of PF2 is at point Q, Q is the midpoint between point P and point F2, OQ is the radius of the circle, OQ is perpendicular to PF2, OQ//PF 1.

So PF 1=2OQ=2b, and PF 1 is perpendicular to PF2.

And PF 1+PF2=2a.

So | pf1| 2+| pf2 | 2 = | f1F2 | 2.

That is 4b 2+(2a-2b) 2 = 4c 2.

Derive 3b=2a.

According to a 2-b 2 = c 2

The eccentricity e=c/a=√5/3 is deduced.