解：(1)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=1y=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...
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