Question

Find an equation in standard form for the hyperbola with vertices at (0, ±8) and asymptotes at y = (± 4/3)x.

a) 9y^2 - 16x^2 = 144
b) 16y^2 - 9x^2 = 144
c) 25y^2 - 16x^2 = 144
d) 16y^2 - 25x^2 = 144

Answer 1

The equation in standard form for the hyperbola is 9x^2 - 16y^2 = 144.

Step-by-step explanation:

The equation in standard form for the hyperbola can be found using the following steps:

1. The center of the hyperbola is at the origin, so the equation starts as x^2/a^2 - y^2/b^2 = 1.
2. The vertices are given as (0, ±8), so the value of b is 8.
3. The asymptotes are given as y = (±4/3)x, so the value of a is 4/3.
4. Plugging in the values of a and b, the equation becomes 9x^2 - 16y^2 = 144.

Therefore, the correct equation in standard form for the hyperbola is 9x^2 - 16y^2 = 144 (option a).