Answer
See graph. Foci $(0,\pm2\sqrt 5)$; asymptotes $y= \pm\frac{1}{2}x$
Work Step by Step
Step 1. Rewrite the given equation as $-\frac{x^2}{16}+\frac{y^2}{4}=1$. We have $a=2, b=4$ and thus $c=\sqrt {4^2+2^2}=2\sqrt 5$. The foci are on the y-axis.
Step 2. See graph; we can locate the foci at $(0,\pm2\sqrt 5)$
Step 3. The asymptotes can be found as $y=\pm\frac{2}{4}x=\pm\frac{1}{2}x$
