Answer
Foci: $(-1,2\pm\sqrt{13})$
Vertices: $(-1,0)$ and $(-1,4)$
Asymptotes: $y=-\frac{2}{3}x+\frac{4}{3}$ and $y=\frac{2}{3}x+\frac{8}{3}$
Work Step by Step
Recall: The hyperbola $\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$ has the foci $(h, k\pm c)$ where $c^2=a^2+b^2$, the vertices $(h,k\pm a)$, and the asymptotes $y-k=\pm(a/b) (x-h)$.
We have $9y^2-4x^2-36y-8x=4$.
Rewrite this equation:
$9y^2-36y-4x^2-8x=4$
$9y^2-36y+36-4x^2-8x-4=4+36-4$
$9(y-2)^2-4(x+1)^2=36$
$\frac{(y-2)^2}{4}-\frac{(x+1)^2}{9}=1$
Then,
$h=-1$
$k=2$
$a^2=4\to a=2$
$b^2=9\to b=3$
$c=\sqrt{a^2+b^2}=\sqrt{4+9}=\sqrt{13}$
$a/b=2/3$
Find the foci:
$(h,k\pm c)=(-1,2\pm \sqrt{13})$
Find the vertices:
$(h,k\pm a)=(-1,2\pm 2)$
$(-1,0)$ or $(-1,4)$
Find the asymptotes:
$y-k=\pm (a/b)(x-h)$
$y-2=\pm 2/3(x+1)$
$y=\pm 2/3(x+1)+2$
$y=-2/3(x+1)+2$ or $y=2/3(x+1)+2$
$y=-\frac{2}{3}x+\frac{4}{3}$ or $y=\frac{2}{3}x+\frac{8}{3}$
