由题意得：右准线a^2/c=2 (a>0,c>0)
线段PF1中点坐标为((2-c)/2,3^0.5/2),设为点Q
由PF1·QF2=0得,3c^2+4c-7=0
∴c=1,a=2^0.5,b=1
椭圆方程为x^2/2+y^2=1
(2)设Q点(xq,yq)
x^2/2+y^2=1
y=kx+m
联立得(2k^2+1)x^2+4kmx+2m^2-2=0
△>0
向量OA+向量OB=(xa+xb,ya+yb)
(xa+xb,ya+yb)=(-4km/(2k^2+1),2m/(2k^2+1))
要使得向量OA+向量OB=x向量OQ
则(-4km/(2k^2+1),2m/(2k^2+1))=X(xq,yq)
∵Q点在椭圆上
∴点(-4km/(2k^2+1)X,2m/(2k^2+1)X)也在椭圆上,代入椭圆方程得
X^2=4m^2/(2k^2+1)
又∵△>0
∴2k^2+1>m^2
代入X^2,解得X∈(-2,2)