Answer
Asymptotes: $y=\displaystyle \pm\frac{2}{3}x$
Foci: $(0,2\sqrt{13})$ and $(0, -2\sqrt{13})$.
Work Step by Step
The hyperbola $\displaystyle \frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$
has a vertical transverse axis and two asymptotes
$y=\displaystyle \pm\frac{a}{b}x$
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$a=4, \quad b=6$
For the foci,
$c^{2}=a^{2}+b^{2}=16+36=52$
$c=\sqrt{52}=2\sqrt{13}\approx 7.211$
Vertices: $(0,4)$ and $(0, -4)$
Asymptotes:
$y=\displaystyle \pm\frac{a}{b}x=\pm\frac{4}{6}x=\pm\frac{2}{3}x$
Foci: $(0,2\sqrt{13})$ and $(0, -2\sqrt{13})$.
