Answer
Equation: $\frac{y^2}{4}-\frac{x^2}{8}=1$
Foci: $(0,\pm2\sqrt{3})$
Asymptotes: $y=\pm\frac{1}{\sqrt{2}}x$
Work Step by Step
Recall: The hyperbola whose the vertices are $(0,\pm a)$ has the equation $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$.
In the figure, the vertices are $(0,\pm2)$.
We get $a=2$ and so $a^2=4$.
The hyperbola also passes through the point $(8,6)$.
Then,
$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\frac{6^2}{4}-\frac{8^2}{b^2}=1$
$\frac{36}{4}-\frac{64}{b^2}=1$
$9-\frac{64}{b^2}=1$
$\frac{64}{b^2}=8$
$b^2=8$
So, the equation of the hyperbola is $\frac{y^2}{4}-\frac{x^2}{8}=1$.
Find the value of $c$:
$c=\sqrt{a^2+b^2}=\sqrt{4+8}=\sqrt{12}=2\sqrt{3}$
Find the foci:
$(0,\pm c)=(0,\pm2\sqrt{3})$
Find the asymptotes:
$y=\pm (a/b)x$
$y=\pm (2/\sqrt{8})x$
$y=\pm 1/\sqrt{2}x$
