91Ó°ÊÓ

The equation of the hyperbola is

$\frac{{\left(y+3\right)}^2}{4}-\frac{{\left(x-2\right)}^2}{9}=1$

The hyperbola with center $\left(h,k\right)$and transverse axis parallel to $y$-axis is given by,

role="math" localid="1647139124552" $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$; $b^2=c^2-a^2$

where foci are $\left(h,kÂ\pm c\right)$, vertices are $\left(h,kÂ\pm a\right)$and asymptotes are$y-k=Â\pm \frac{a}{b}\left(x-h\right)$.

The given hyperbola is

$\frac{{\left(y+3\right)}^2}{4}-\frac{{\left(x-2\right)}^2}{9}=1$

It can be written as,

$\frac{{\left(y+3\right)}^2}{2^2}-\frac{{\left(x-2\right)}^2}{3^2}=1$

This is standard form of hyperbola $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$.

So, $a=2,b=3$

$h=2,k=-3$

Since $c^2=a^2+b^2$,

$c^2=4+9$

$c^2=13$

$c=\sqrt{13}$

So, The center of the hyperbola is $\left(h,k\right)≡\left(2,-3\right)$,

The foci are $\left(h,kÂ\pm c\right)≡\left(2,-3Â\pm \sqrt{13}\right)$,

The vertices are role="math" localid="1647141442843" $\left(h,k+a\right)≡\left(2,-3+2\right)≡\left(2,-1\right)$and

$\left(h,k-a\right)≡\left(2,-3-2\right)≡\left(2,-5\right)$

Since foci and vertices all lie on the line $x=2$, the transverse axis is parallel to $y$-axis.

The asymptotes of the hyperbola is

$y+3=Â\pm \frac{2}{3}\left(x-2\right)$

The graph of the hyperbola$\frac{{\left(y+3\right)}^2}{4}-\frac{{\left(x-2\right)}^2}{9}=1$ is