Find an equation in standard form for the hyperbola with vertices at (0, ±6) and asymptotes at y = ± 3/4x

Equation is y²/36 - x²/64 = 1

Step-by-step explanation:

It is given that vertices are (0, ±6)  that makes this a vertical aligned hyperbola.The center of the hyperbola is midway between the left and right vertex which makes the center equal to (0,0).The general equation of vertical hyperbola is:

y²/a² - x²/b² = 1

where a is the distance between the vertex and the center,which is equal to a=6.The equation for the asymptotes of a vertical aligned hyperbola is:

y = +/- a/b x

It is given that:

a/b = 3/4

6/b = 3/4

By cross multiplication:

6*4 = 3*b

24= 3b

Divide both sides by 3

24/3 = 3b/3

b = 8

The formula for this vertically aligned hyperbola is:

y²/36 - x²/64 = 1

Equation is y²/36 - x²/64 = 1

1/8 3/16 3/4

step-by-step explanation:

you put them as all the same denominator and multiply