91Ó°ÊÓ

The given function is

$y^2-4x^2-4y-8x-4=0$

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

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

where foci at $\left(h,kÂ\pm c\right)$, vertices at $\left(h,kÂ\pm a\right)$and asymptotes are

$y-k=Â\pm \frac{a}{b}\left(x-h\right)$

The given function is

$y^2-4x^2-4y-8x-4=0$

Convert into standard form of hyperbolas.

Grouping terms,

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

Complete squaring,

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

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

Divide by $4$,

role="math" localid="1647361732237" $\frac{{\left(y-2\right)}^2}{4}-{\left(x+1\right)}^2=1$

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

This is the standard form og hyperbola.

So, $a=2,b=1$

$h=-1,k=2$

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

$c^2=4+1$

$c^2=5$

$c=\sqrt{5}$

Thus, the center of the hyperbola is $\left(h,k\right)≡\left(-1,2\right)$,

The foci at $\left(h,kÂ\pm c\right)≡\left(-1,2Â\pm \sqrt{5}\right)$,

The vertices at $\left(h,k+a\right)≡\left(-1,2+2\right)≡\left(-1,4\right)$and

$\left(h,k-a\right)≡\left(-1,2-2\right)≡\left(-1,0\right)$

Since foci and vertices all lie on the line $x=-1$, transverse axis is parallel to $y$-axis.

The asymptotes are

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

So,$y-2=Â\pm 2\left(x+1\right)$

The graph of the hyperbola is