Answer
See graph
Work Step by Step
We are given the hyperbola:
$(x+3)^2-9(y-4)^2=9$
Bring the equation to the standard form:
$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$
$\dfrac{(x+3)^2}{9}-\dfrac{9(y-4)^2}{9}=1$
$\dfrac{(x+3)^2}{9}-\dfrac{(y-4)^2}{1}=1$
The transverse axis is parallel to the $x$-axis.
Determine $h,k,a,b,c$:
$h=-3$
$k=4$
$a^2=9\Rightarrow a=\sqrt 9=3$
$b^2=1\Rightarrow b=\sqrt 1=1$
$c^2=a^2+b^2$
$c^2=9+1$
$c^2=10$
$c=\sqrt {10}$
The centre of the hyperbola is:
$(h,k)=(-3,4)$
Determine the coordinates of the vertices:
$(h-a,k)=(-3-3,4)=(-6,4)$
$(h+a,k)=(-3+3,4)=(0,4)$
Determine the coordinates of the foci:
$(h-c,k)=(-3-\sqrt{10},4)$
$(h+c,k)=(-3+\sqrt{10},4)$
Determine the asymptotes:
$y-k=\pm\dfrac{b}{a}(x-h)$
$y-4=\pm\dfrac{1}{3}(x+3)$
Graph the hyperbola:
