1. Find the coordinates of point P: b 2x 2-2pa 2x-(ab) 2 = 0...( 1); (p/2)=c，p=2c.....(2)
△=(-2pa^2)^2-4b^2[-(ab)^2]=4a^2[(pa)^2+b^4]=4a^2[4(a^2+b^2)a^2+b^4]
=[2a(2a^2+b^2)]^2；
X 1, 2 = {2pa 2+/-2a (2a 2+b 2)}/(2b 2) (take a positive number and discard a negative number).
px=[pa^2+a(2a^2+b^2)]/b^2=[2ca^2+a(2a^2+b^2)]/b^2；
py =+/-√( 2px)=+/-2√CX； Get the coordinates of point p: (Px, Py)
2. Find E: pf1= √ [(c-px) 2+(0-py) 2] = √ [(c-px) 2+(4cpx)] ... (3)
According to the meaning of the question:1/cos2 ∠ pf1F2 = (pf1) 2 =1+(4cpx)/(c-px) 2 = (7/5).
4cpx*25=24(c^2-2cpx+px^2)； 6px^2-37cpx+6c^2=(px-6c)(6px-c)=0
Px 1=6c，Px2 = c/6；
6cb^2=2ca^2+a(2a^2+b^2)； 2c(3b^2-a^2)=a(2a^2+b^2)....(4);
De: e = c/a = (2a 2+b 2)/(6b 2-2a 2) = 2，3，√ 2，√3； (After verification, there is no qualified answer. )
3. Calculate the value of e: e (6b2-2a2) = 2a2+b2; (6e- 1)b^2=(2e+2)a^2； b^2=(2e+2)a^2/(6e- 1)；
a^2+b^2=a^2[ 1+(2e+2)/(6e- 1)a^2； Divide both sides of the equation by 2 at the same time, and you get:
e^2=(a^2+b^2)/a^2=(8e+ 1)/(6e- 1)； e^2(6e- 1)=8e+ 1； Namely: 6E3-E2-8e-1= 0;
6e^3-e^2-7e-(e+ 1)=e(6e^2-e-7)-(x+ 1)=e(6e-7)(e+ 1)+(e+ 1)=(e+ 1)(6e^2-7e- 1)=0；
E=- 1 (unreasonable, discarded); Then: (6E2-7E-1) = 0; E=(7+/-√73)/ 12 because of e >；; 0，e =(7+√73)/ 12；
cb^2/6=2ca^2+a(2a^2+b^2)； c(b^2- 12a^2)=6a(2a^2+b^2)；
e=c/a=6(2a^2+b^2)/(b^2- 12a^2)； Yes: e (B2-12a2) =12a2+6b2;
(e-6)b^2= 12(e+ 1)a^2； c^2=a^2+b^2=[ 1+ 12(e+ 1)/(e-6)]a^2=( 13e+6)a^2/(e-6)
e^2(e-6)-( 13e+6)=e^3-6e^2- 13e-6=e(e-7)(e+ 1)-(e+ 1)=(e+ 1)(e^2-7e- 1)=0；
Similarly: e = (7+/-4 √ 3)/2; e=(7+4√3)/2 .
After calculation, there is no qualified answer.
I searched for a long time and found no mistakes in solving the problem. Errors in calculation are not excluded; But I really can't find anything wrong. Please check it again.