find an hyperbola equation with foci (0, +/-8) and asymptotes; y=+/-1/2x?

Answer 1

I use a slightly different approach to conics than the one Steve just showed you in your previous post, but the end results are the same.
I keep the values of a and b always associated with the x and y, no matter which way the hyperbola is oriented.
If it has its main axis horizontally, I use
x^2/a^2 - y^2/b^2 = 1
If it has its main axis vertically, I use
x^2/a^2 - y^2/b^2 = -1

I also sketch the box formed by the asymptotes which forms the skeleton for my graph

from y = (1/2)x I know b/a = 1/2
a = 2b

and in any hyperbola: a^2 + b^2 = c^2
(2b)^2 + b^2 = 64
5b^2 = 64
b^2 = 64/5
b = 8/√5
a = 16/√5 ----> a^2 = 256/5

equation:
x^2/(256/5) - y^2/(64/5) = -1
or
5x^2/256 - 5y^2/64 = -1