Answer
Center: $(3,-2)$
Foci: $(3,-6),(3,2)$
Vertices: $(3,-2-3\sqrt 2),(3,-2+3\sqrt 2)$
See graph
Work Step by Step
We are given the ellipse:
$9(x-3)^2+(y+2)^2=18$
Put the equation in standard form:
$\dfrac{9(x-3)^2}{18}+\dfrac{(y+2)^2}{18}=1$
$\dfrac{(x-3)^2}{2}+\dfrac{(y+2)^2}{18}=1$
The equation is in the form:
$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$
Identify $h,k,a,b$:
$h=3$
$k=-2$
$a^2=18\Rightarrow a=\sqrt{18}=3\sqrt 2$
$b^2=2\Rightarrow b=\sqrt 2$
Determine $c$:
$a^2=b^2+c^2$
$c^2=a^2-b^2$
$c^2=18-2$
$c^2=16$
$c=\sqrt {16}=4$
Determine the center:
$(h,k)=(3,-2)$
Determine the foci:
$(h,k-c)=(3,-2-4)=(3,-6)$
$(h,k+c)=(3,-2+4)=(3,2)$
Determine the vertices:
$(h,k-a)=(3,-2-3\sqrt 2)$
$(h,k+a)=(3,-2+3\sqrt 2)$
Determine the $x$-intercepts:
$(h-b,k)=(3-\sqrt 2,-2)$
$(h+b,k)=(3+\sqrt 2,-2)$
Graph the ellipse:
