Posted by Paragon on .
Find the equation of a hyperbola with foci of
(0,8), and (0,8)
and Asymptotes of
y=4x and y=4x
Please show work!!!

Adv. Algebra Conics 
Reiny,
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 
Adv. Algebra Conics 
Reiny,
correction:
using x^2/a^2 = y^2/b^2 = 1 should say
using x^2/a^2  y^2/b^2 = 1 
Adv. Algebra Conics 
Paragon,
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 
Adv. Algebra Conics 
Reiny,
<..but wouldn't 32^2 be 1024>
of course, good for you for catching that.
let's blame it on a "senior moment"