由題意得：右準(zhǔn)線a^2/c=2 (a>0,c>0)
線段PF1中點(diǎn)坐標(biāo)為((2-c)/2,3^0.5/2),設(shè)為點(diǎn)Q
由PF1·QF2=0得,3c^2+4c-7=0
∴c=1,a=2^0.5,b=1
橢圓方程為x^2/2+y^2=1
(2)設(shè)Q點(diǎn)(xq,yq)
x^2/2+y^2=1
y=kx+m
聯(lián)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
則(-4km/(2k^2+1),2m/(2k^2+1))=X(xq,yq)
∵Q點(diǎn)在橢圓上
∴點(diǎn)(-4km/(2k^2+1)X,2m/(2k^2+1)X)也在橢圓上,代入橢圓方程得
X^2=4m^2/(2k^2+1)
又∵△>0
∴2k^2+1>m^2
代入X^2,解得X∈(-2,2)