由e=√3/2得a=2b；
设椭圆方程为x^2/4+y^2=b^2
将直线方程与椭圆方程联立得
x^2+2x+2-2b^2=0
x1=-1+√(2b^2-1)、x2=-1-√(2b^2-1)
OM=(1/2)OA+(√3/2)OB=(x1/2+√3x2/2,y1/2+√3y2/2)
xM=x1/2+√3x2/2=-(1+√3)/2+(1-√3)√(2b^2-1)/2
yM=(x1/2+1)/2+√3(x2/2+1)/2=(1+√3)/2+xM/2
M在椭圆上,所以xM^2+4yM^2=4b^2
xM^2+(1+√3)xM+1+√3/2-2b^2=0
(1+√3)^2/4+(1-√3)^2(2b^2-1)/4+√(2b^2-1)-(1+√3)^2/4-√(2b^2-1)+1+√3/2-2b^2=0
即√3-√3b^2=0
b^2=1
所以椭圆方程为x^2/4+y^2=1.