Answer
center $(0,0)$, transverse axis $y$-axis, vertices $(0,-4),(0,4)$, foci $(0,-2\sqrt {5}),(0,2\sqrt {5})$, and asymptotes $y =\pm2x$.
See graph.
Work Step by Step
1. Given $\frac{y^2}{16}-\frac{x^2}{4}=1$, we can find the center $(0,0)$, transverse axis $y$-axis, $a=4,b=2$, vertices $(0,-4),(0,4)$, $c=\sqrt {a^2+b^2}=\sqrt {16+4}=2\sqrt {5}$, foci $(0,-2\sqrt {5}),(0,2\sqrt {5})$, and asymptotes $y=\pm\frac{a}{b}x=\pm2x$.
2. See graph.
