91Ó°ÊÓ

The equation of the hyperbola is

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

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

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

where foci are$\left(h,kÂ\pm c\right)$, vertices are$\left(h,kÂ\pm a\right)$and asymptotes arerole="math" localid="1647140442921" $y-k=Â\pm \frac{a}{b}\left(x-h\right)$.

The given hyperbola is

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

It can be written as,

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

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

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

So, $a=2,b=1$

$h=-2,k=2$

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

$c^2=4+1$

$c^2=5$

So, $c=\sqrt{5}$

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

The foci are $\left(h,kÂ\pm c\right)≡\left(-2,2Â\pm \sqrt{5}\right)$,

The vertices are $\left(h,k+a\right)≡\left(-2,2+2\right)≡\left(-2,4\right)$and

$\left(h,k-a\right)≡\left(-2,2-2\right)≡\left(-2,0\right)$

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

The asymptotes of the hyperbola is

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

So, role="math" localid="1647140887873" $y-2=Â\pm 2\left(x+2\right)$

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