Answer
center $(0,0)$, transverse axis $x$-axis, vertices $(-\sqrt 2,0),(\sqrt 2,0)$, foci $(-\sqrt {6},0),(\sqrt {6},0)$, and asymptotes $y =\pm\sqrt 2 x$.
See graph.
Work Step by Step
1. Given $2x^2-y^2=4\Longrightarrow \frac{x^2}{2}-\frac{y^2}{4}=1$, we can find the center $(0,0)$, transverse axis $x$-axis, $a=\sqrt 2,b=2$, vertices $(-\sqrt 2,0),(\sqrt 2,0)$, $c=\sqrt {a^2+b^2}=\sqrt {2+4}=\sqrt {6}$, foci $(-\sqrt {6},0),(\sqrt {6},0)$, and asymptotes $y=\pm\frac{b}{a}x=\pm\sqrt 2 x$.
2. See graph.
