Answer
Vertices: $(0,-4)$ and $(0,4)$
Foci: $(-\sqrt 17,0)$ and $(\sqrt 17,0)$.
Asymptotes: $y = \pm \frac{1}{4} x$
Work Step by Step
Divide both sides of the equation by 16 to put the equation in standard form.
$\frac{y^{2}}{16}-\frac{x^{2}}{1}=1$
The following is a horizontal hyperbola because the $y^{2}$ term is first, while a vertical hyperbola has the $x^{2}$ term first.
$\frac{y^{2}}{16}-\frac{x^{2}}{1}=1$
General equation of a vertical hyperbola:
$\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$ where $a$ is the distance from the center to the vertices.
$a^{2} = 16,$ so $a = 4$, and because the hyperbola is centered at the origin, then the verticles are $(0,-4)$ and $(0,4)$
The distance from the center to the foci is equal to $c$ where $c^{2} = a^{2} +b^{2}$
Plugging in values, we get $c^{2} = 16 +1$ and $c = \sqrt 17$
Thus, the foci are $(-\sqrt 17,0)$ and $(\sqrt 17,0)$.
The slope of the asymptotes passing through the center of the hyperbola is $\pm \frac{a}{b}$
Plugging in values, we get $m = \frac{4}{1}$ so the equations of the asymptotes are $y = \pm 4 x$
