设交点为A(x1,y1),B(x2,y2)
设过点M(m,0)的直线为y=-√3/3*(x-m)
带入圆方程得
x^2+(x-m)^2/3=1,整理得
4x^2-2mx+m^2-3=0,由韦达定理有
x1+x2=m/2, x1x2=(m^2-3)/4;y1+y2=-(x1+x2-2m)/√3=√3/2*m
y1y2=(x1-m)(x2-m)/3=[x1x2-m(x1+x2)+m^2]/3=(m^2-1)/4
又向量AM=(m-x1,-y1), 向量MB=(x2-m,y2)
且向量AM=2向量MB
∴有m-x1=2(x2-m), -y1=2y2
与上述韦达定理所得等式联立,可解得
x1=-2m, x2=5m/2
y1=√3*m, y2=-√3/2*m
m^2=1/7, m=±√(1/7)
∴m的取值为m=±√(1/7)