Answer
Vertices: $(0,\pm 4)$
Foci: $(0,\pm\sqrt{17})$.
Asymptotes: $y = \pm 4x$
Work Step by Step
Recall: The hyperbola $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ has the foci $(0,\pm c)$ where $c^2=a^2+b^2$, the vertices $(0,\pm a)$, and the asymptotes $y=\pm (a/b)x$.
We have $y^2-16x^2=16$.
It can be rewritten as $\frac{y^2}{16}-\frac{x^2}{1}=1$.
Then,
$a^2=16\to a=4$
$b^2=1\to b=1$
$c=\sqrt{a^2+b^2}=\sqrt{16+1}=\sqrt{17}$
$a/b=4/1=4$
So, the foci are $(0,\pm\sqrt{17})$, the vertices are $(0,\pm 4)$, and tbe asymptotes are $y=\pm 4x$.
