Answer
See graph
Work Step by Step
We are given the hyperbola:
$(x-3)^2-4(y+3)^2=4$
Bring the equation to the standard form:
$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$
$\dfrac{(x-3)^2}{4}-\dfrac{4(y+3)^2}{4}=1$
$\dfrac{(x-3)^2}{4}-\dfrac{(y+3)^2}{1}=1$
The transverse axis is parallel to the $x$-axis.
Determine $h,k,a,b,c$:
$h=3$
$k=-3$
$a^2=4\Rightarrow a=\sqrt 4=2$
$b^2=1\Rightarrow b=\sqrt 1=1$
$c^2=a^2+b^2$
$c^2=4+1$
$c^2=5$
$c=\sqrt 5$
The centre of the hyperbola is:
$(h,k)=(3,-3)$
Determine the coordinates of the vertices:
$(h-a,k)=(3-2,-3)=(1,-3)$
$(h+a,k)=(3+2,-3)=(5,-3)$
Determine the coordinates of the foci:
$(h-c,k)=(3-\sqrt{5},-3)$
$(h+c,k)=(3+\sqrt{5},-3)$
Determine the asymptotes:
$y-k=\pm\dfrac{b}{a}(x-h)$
$y+3=\pm\dfrac{1}{2}(x-3)$
Graph the hyperbola:
