⑴设Q(x0,0),
F(-c,0)A(0,b),FA=(c,b),AQ=(x0,-b)
∵ FA⊥AQ,
∴ cx0-b2=0,x0=b2/c
P(x1,y1),
AP=8/5PQ
x1=8b2/13c,y1=5/13b
P在椭圆上
(8b2/13c)2/a2+(5/13b)2/b2=1
2b2=3ac,
⑵(a2-c2)=3ac,2e2+3e-2=0,
e= 1/2.
(2)F(- 1/2a,0)Q (3/2a,0),
圆心( 1/2a,0),r= 1/2|FQ|=a
|1/2a+3|/2=a,
a=2,
∴c=1,b= √3,
x2/4+y2/3=1