设P点坐标(-c,y0),左准线方程为:x=-a/e,x=-a^2/c,
P至左准线距离|PQ|=a^2/c-c=4-c,根据圆锥曲线定义,y0/|PQ|=e=c/a,
y0=c(4-c)/a,(设y0>0)
OP直线斜率k1=y0/(-c)=-yo/c=-(4-c)/a,
AB直线斜率k2=(-b)/[0-(-a)]=-b/a,
∵AB‖OP,
∴k1=k2,-(4-c)/a=-b/a,b=4-c,.(1),
|-a^2/c|=4,a^2=4c,.(2),
从(1)式得:√(a^2-c^2)=4-c,a^2-c^2=16-8c+c^2,
c^2-6c+8=0,
(c-2)(c-4)=0,
c=2或c=4,但当c=4时,a=4,因a>c,故不符合题意,应舍去)
代入(2)式,
a=2√2,b=2,
∴方程为:x^2/8+y^2/4=1.