I'll tell you the rule first:
The general form of the hyperbola having a vertical axis and centered at the origin is:
(y/a)^2 ‑ (x/b)^2 = 1
Asymptotes:
y= ±(a/b)x
Or (y/a)^2 - (x/b)^2 = 0
So in your problem the asymptotes are:
2y ± 6x = 0
2y = ±6x
y = ±(3/2)x
So, a=3 b=2
The equations are:
(y/3)^2 - (x/2)^2 = 1
(y^2)/8 - (x^2)/4 = 1
Hope it helps you.
