Solution:

( 1) a=√ 3，c=2。 The focus is on the x axis.

Then b 2 = c 2-a 2 = 1

So the hyperbolic equation is x 2/3-y 2 =1.

(2)

The slope of NM =k, then the slope of the median vertical line =-1/k.

Let n and m be (x 1, y 1), (x2, y2) and let the midpoint of NM be (a, b).

Let the bisector be L: Y =-X/K+B2.

Because l passes through (0,-1), b2=- 1.

L is y=-x/k- 1.

Because (x12-x2 2)/3 = (y12+1)-(y2 2+1)

(x 1+x2)/3(y 1+y2)=(y 1-y2)/(x 1-x2)= k

Then a/3b=k,

And point o is also on the straight line l, then b=-a/k- 1 (substitute k=a/3b).

Get b=- 1/4 and k=-4a/3.

Obviously, point o is also on the straight line y=kx+m, so b = ka+m.

Then-1/4 =-3k2/4+m.

3k^2=4m+ 1

Substitute y=kx+m into hyperbolic equation and eliminate y.

X 2/3-k 2x 2-2kmx-m 2-1= 0 If the equation has two real roots.

Then 4m2k2-4 (-m2-1) (1/3-k2) > 0.

m^2/3-k^2+ 1/3>； 0

m^2+ 1>； 3k^2=4m+ 1

Solving m>4 or m