已知双曲线x2/3-y2=1,直线y=kx+m（k、m≠0）与双曲线交于C、D两点,且CD的垂直平分线过点B（0,-1）,试求m的取值范围.
CD的斜率=k,则垂直平分线的斜率=-1/k
设C、D两点为（x1,y1),(x2,y2),设CD中点M为(a,b),
可设平分线为L：y=-x/k+b2
因L经过（0,-1）得b2=-1
L为y=-x/k-1
因(x12-x22)/3=(y12+1)-(y22+1)
=>(x1+x2)/3(y1+y2)=(y1-y2)/(x1-x2)=k
则a/3b=k,
又M点也在直线L上则b=-a/k-1(将k=a/3b代入）
得b=-1/4,k=-4a/3
显然M点也在直线y=kx+m上,则b=ka+m
则-1/4=-3k2/4+m
3k2=4m+1
将y=kx+m代入双曲线方程消去y
x2/3-k2x2-2kmx-m2-1=0要使方程有两实根
则4m2k2-4(-m2-1)(1/3-k2)>0
=>m2/3-k2+1/3>0
=>m2+1>3k2=4m+1
解得m>4或m