hyperbola with vertices at (0,+/-2), and asymptotes y=+- 1/2x

Answer 1

since the vertices are on the y-axis, we will have

y^2/a^2 - x^2/b^2 = 1
the slope of the asymptotes is b/a, so

y^2 - x^2/4 = 1

but that has vertices at (0,+/-1) so

y^2/4 - x^2/16 = 1

To verify, see

http://www.wolframalpha.com/input/?i=hyperbola+y%5E2%2F4+-+x%5E2%2F16+%3D+1