According to the definition of ellipse, when point P is at the right vertex of the long axis (X axis), |PF 1| is the largest, which is equal to the length of the long semi-axis plus the half focal length, that is, A+C = A+√ (A 2-B 2).

When point P is at the left vertex of the long axis (X axis), |PF 1| is the smallest, which is equal to half the focal length of the long semi-axis.

That is, a-c = a-√ (a 2-b 2).

Let p (x0, y0), c = √ (a 2-b 2) = 2 √ 5, focus f1(-2 √ 5,0), F2 (2 √ 5,0),

PF 1⊥PF2, pf1linear slope k 1=y0/(x0+2√5), PF2 linear slope k2=y0/(x0-2√5),

K 1 =- 1/k2, y0/(x0+2 √ 5) =-(x0-2 √ 5)/y0, x0 2+y0 2 = 20, and the point p is on the ellipse.

x0^2/25+y0^2/5= 1,y0=√5/2 x0 = 5√3/2

* * * Four points meet the conditions, x0=5√3/2, Y0 = √ 5/2;

x0=5√3/2，y0 =-√5/2；

x0=-5√3/2，y0 =√5/2；

x0=-5√3/2，y0 =-√5/2；