## Precalculus (6th Edition) Blitzer

$\dfrac{x^2}{16}-\dfrac{y^2}{64}=1$ With horizontal traverse axis.
Endpoints: $(-4,0), (4, 0)$ and Asymptote: $y=2x$ We are given that $a=4$ Since $\dfrac{b}{a}=2 \implies b=8$ so, $b^2=64$ Thus, the equation for hyperbola is: $\dfrac{x^2}{16}-\dfrac{y^2}{64}=1$ And the traverse axis is horizontal.