WILL MARK BRAINLIEST!! Find an equation in standard form for the hyperbola with vertices at (0, ±8) and asymptotes at y = ±4 divided by 3.x. y squared over 9 minus x squared over 16 = 1 y squared over 36 minus x squared over 64 = 1 y squared over 64 minus x squared over 9 = 1 y squared over 64 minus x squared over 36 = 1

y^2 / 64 - x^2 /36 =1

Step-by-step explanation:

hyperbola with vertices at (0, ±8)

asymptotes at y = ±4/3x

The general form is

y^2 /a^2 - x^2/ b^2 =1

where the asymptote is y = ± a/b x

but we also have The vertex is ( 0 ,a)

rewriting the asymptote as 8/6 so a =8 and b = 6

y^2 / 8^2 -x^2/6^2 = 1

y^2 / 64 - x^2 /36 =1

one that is bigger than 3 1/2