Answer
Center: $\quad \quad (3,-2)$
Foci: $\quad \quad (3,2), \quad (3,-6)$
Vertices: $\quad (3,-2+3\sqrt{2}), \quad (3,-2-3\sqrt{2})$
Work Step by Step
Divide with $18$ to obtain the standard form
$\displaystyle \begin{aligned}\frac{9(x-3)^{2}}{18}+\frac{(y+2)^{2}}{18}&=1\displaystyle \\\frac{(x-3)^{2}}{2}+\frac{(y+2)^{2}}{18}&=1\displaystyle \end{aligned}$
1. $ \quad $The form of the equation is (See table 3)
Major axis vertical (parallel to the y-axis)
$\begin{array}{lll}
\text{ Foci}&\text{ Vertices}&\text{ Equation}\\
{(h,k+c)}&{(h,k+a)}&{\displaystyle \frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1}\\
{(h,k-c)}&{(h,k-a)}&{a \gt b \gt 0\text{ and }b^{2}=a^{2}-c^{2}}\end{array}$
2. $ \quad $Center = $(h,k)=(3,-2)$,
3.$ \quad a=\sqrt{18}=3\sqrt{2},\quad \quad b=\sqrt{2}$
$(a\approx 4.243 \quad \quad b\approx 1.414 )$
Find $c: $
$c^{2}=a^{2}-b^{2}=18-2=16$
$c=4$
Vertices: $ \quad (h,k\pm a)=(3,-2\pm 3\sqrt{2})$
Foci: $ \quad (h,k\pm c) = (2,3\pm 4)$
$(3,2), \quad (3,-6)$
