Answer
See graph
Foci: $(-3,-2),(7,-2)$
Work Step by Step
We are given the hyperbola:
$\dfrac{(x-2)^2}{9}-\dfrac{(y+2)^2}{16}=1$
The equation is in the standard form:
$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$
The transverse axis is parallel to the $x$-axis.
Determine $h,k,a,b,c$:
$h=2$
$k=-2$
$a^2=9\Rightarrow a=\sqrt {9}=3$
$b^2=16\Rightarrow b=\sqrt {16}=4$
$c^2=a^2+b^2$
$c^2=9+16$
$c^2=25$
$c=\sqrt{25}$
$c=5$
The centre of the hyperbola is:
$(h,k)=(2,-2)$
Determine the coordinates of the vertices:
$(h-a,k)=(2-3,-2)=(-1,-2)$
$(h+a,k)=(2+3,-2)=(5,-2)$
Determine the coordinates of the foci:
$(h-c,k)=(2-5,-2)=(-3,-2)$
$(h+c,k)=(2+5,-2)=(7,-2)$
Determine the asymptotes:
$y-k=\pm\dfrac{b}{a}(x-h)$
$y+2=\pm\dfrac{4}{3}(x-2)$
Graph the hyperbola:
