Question

Find an equation in standard form for the hyperbola with vertices at (0, ±8) and asymptotes at y=+-1/2x

Answer 1

\[\frac{ y^2 }{ 256 }-\frac{ x^2 }{ 64 }=1\]
\[\frac{ y^2 }{ 64 }-\frac{ x^2 }{ 256 }=1\]
\[\frac{ y^2 }{ 64 }-\frac{ x^2 }{ 16 }=1\]
\[\frac{ y^2 }{ 64 }-\frac{ x^2 }{ 64}=1\]