Question

Find an equation for the hyperbola that satisfies the given conditions.
Foci (0, ±8), vertices (0, ±6)

Answer 1

well
the distance from the origin to the focus(which is (f)) is equal to the distance from the origin to vertex(which is (a)) times the value (e) where (e) is the eccentricity
so
f=a*e , e=(f/a) , e=8/6=(4/3)
from properties of hyperbola
f^2=a^2+b^2 , b^2=f^2-a^2 , b^2=64-36=28 ,b=+-(28)^(1/2)
now you have a and b
you can substitute in the equation of the hyperbola
((x^2)/a^2)-((y^2)/b^2)=1