Answer
asymptotes: $y=\pm\frac{2}{3}x$
foci: $(0, \pm2\sqrt {13})$
Work Step by Step
Step 1. From the given equation $\frac{y^2}{16}-\frac{x^2}{36}=1$, we have $a=4, b=6, c=\sqrt {a^2+b^2}=2\sqrt {13}$ centered at $(0,0)$, with a vertical transverse axis.
Step 2. We can find the vertices as $(0,\pm4)$ and asymptotes as $y=\frac{a}{b}x=\pm\frac{2}{3}x$
Step 3. We can graph the equation as shown in the figure with foci at $(0, \pm2\sqrt {13})$
