Find the equation of a hyperbola with foci of

Question

(0,8), and (0,-8)

and Asymptotes of

y=4x and y=-4x

Please show work!!!

Answer 1

In a hyperbola with centre at the origin the equation of the asymptote is y = ±(b/a)x.
so b/a = 4 or b = 4a
Also c = 8
then in a^2 + b^2 = c^2 for a hyperbola
a^2 + 16a^2 = 64
17a^2 = 64
a = 8/√17
then b=32/√17

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

x^2/(64/17) - y^2/(32/17) = -1

17x^2 /64 - 17y^2 /64 = -1

Answer 2

correction:
using x^2/a^2 = y^2/b^2 = -1 should say

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

Answer 3

Wow it's that easy. I was making it harder that it is.
I think that's how you set it up, but wouldn't 32^2 be 1024

Answer 4

<..but wouldn't 32^2 be 1024>

of course, good for you for catching that.

let's blame it on a "senior moment"