Find an equation in standard form for the hyperbola with vertices at (0, ±6 and asymptotes at y = ±three divided by four times x..

The hyperbola has vertices: ( 0, +/- 6 ). It means that a = 6 ( semi-minor axis ).
Also, asymptotes are: y = +/- 3/4 x
The formula is: y = +/- b/ a x
So: 3/4 = b/a
b/6 = 3/4
4 b = 6 * 3
4 b = 18
b = 18 : 4
b = 9/2
The equation of the hyperbola is:
x² / a² - y² / b² = 1
x² / 36 - y² / (9/2)² = 1

x² / 36 - 4y² / 81 = 1