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

Answer:

x² / 36 - 4y² / 81 = 1