According to the meaning of the question, it can be used as a reference picture:

If the line segment of PF 1 is equally divided at the tangent point with circle B, then PF 1 = 2 * √ (C 2-B 2), as can be seen from the figure, PF2=2*OQ=2b.

From PF 1+PF2=2a, we get: 2 * √ (C2-B2)+2b = 2a;

Sort out the above formula (rooting): c 2-b 2 = a 2-2ab+b 2, replace c 2 = a 2-b 2 with: 3b 2-2ab = 0,

Solution: b/a = 2/3;

Centrifugal rate? e=c/a=√[ 1-(b/a)^2]=√( 1-4/9)=√5/3；