Answer
hyperbola, center $(0,0)$, vertices $(-\sqrt 2,0),(\sqrt 2,0)$, foci $(-\sqrt {10},0),(\sqrt {10},0)$, and asymptotes $y=\pm2x$.
Work Step by Step
1. Given $4x^2-y^2=8$ or $\frac{x^2}{2}-\frac{y^2}{8}=1$, we can identify it as a hyperbola with a horizontal transverse axis, $a=\sqrt 2, b=2\sqrt 2$ and $c=\sqrt {a^2+b^2}=\sqrt {10}$.
2. We can find its center $(0,0)$, vertices $(-\sqrt 2,0),(\sqrt 2,0)$, foci $(-\sqrt {10},0),(\sqrt {10},0)$, and asymptotes $y=\pm\frac{b}{a}x=\pm2x$.
