y=kx+1与3x^2-y^2=1联立消去y得(3-k^2)x^2-2kx-2=0.∵直线与曲线有两个交点,∴判别式△＞0,解得：-√6＜k＜√6.（※）由韦达定理,x1+x2=2k/(3-k^2),y1+y2=k(x1+x2)+2=6/(3-k^2),∴A、B的中点坐标（k/(3-k^2),3/(3-k^...