91Ó°ÊÓ

The equation of the hyperbola is

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

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

localid="1647080983389" $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$; $b^2=c^2-a^2$

where foci are 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,

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

It can be written as,

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

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

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

So, localid="1647081047156" $a=3,b=1$

$h=-4,k=3$

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

$c^2=9+1$

$c^2=10$

So,$c=\sqrt{10}$

Thus, the center of the hyperbola is $\left(h,k\right)≡\left(-4,3\right)$,

The foci are $\left(hÂ\pm c,k\right)≡\left(-4Â\pm \sqrt{10},3\right)$,

The vertices are $\left(h+a,k\right)≡\left(-4+3,3\right)≡\left(-1,3\right)$and

$\left(h-a,k\right)≡\left(-4-3,3\right)≡\left(-7,3\right)$

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

The asymptotes of the hyperbola is

$y-3=Â\pm \frac{1}{3}\left(x+4\right)$

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