91Ó°ÊÓ

The equation of the hyperbola is

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

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

role="math" localid="1647080029499" $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$; $b^2=c^2-a^2$

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

The given hyperbola is

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

It cab be written as,

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

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

So, $a=2,b=3$

$h=2,k=-3$

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

$c^2=4+9$

$c^2=13$

So, localid="1647080238000" $c=\sqrt{13}$

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

The focus are $\left(hÂ\pm c,k\right)≡\left(2Â\pm \sqrt{13},-3\right)$,

The vertices are localid="1647080408328" $\left(h+a,k\right)≡\left(2+2,-3\right)≡\left(4,-3\right)$and

$\left(h-a,k\right)≡\left(2-2,-3\right)≡\left(0,-3\right)$

Since all focus and vertices are lie on the line $y=-3$, the transverse axis is parallel to $x$-axis.

The asymptotes of the hyperbola is

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

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