Answer
Center at $\quad(0,0).$
Vertices at $\quad(\pm 3\sqrt{2},0).$
Foci at $\quad(\pm 4,0)$
Work Step by Step
Divide the equation with $18$ ... (we want the standard form)
$\displaystyle \frac{x^{2}}{18}+\frac{9y^{2}}{18}=\frac{18}{18}$
$\displaystyle \frac{x^{2}}{18}+\frac{y^{2}}{2}=1 \qquad$... $\sqrt{18}=3\sqrt{2}\approx 4.243$
$\displaystyle \frac{x^{2}}{(3\sqrt{2})^{2}}+\frac{y^{2}}{(\sqrt{2})^{2}}=1\qquad$... $a=3\sqrt{2},\ b=\sqrt{2}$, main axis: x-axis.
Table 3: Major axis horizontal (parallel to the x-axis)$\begin{array}{lll}
\text{ Foci}&\text{ Vertices}&\text{ Equation}\\
{(h+c, k)}&{(h+a,k)}&{\displaystyle \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1}\\
{(h-c,k)}&{(h-a,k)}&{a \gt b \gt 0\text{ and }b^{2}=a^{2}-c^{2}}\\ \end{array}$
---
The center of the ellipse is at the origin.
The vertices are $(3\sqrt{2},0)$ and $(-3\sqrt{2},0)$ .
Find $c:$
$c^{2}=a^{2}-b^{2}$
$c^{2}=18-2$
$c^{2}=16$
$c=4$
The foci are $(4,0)$ and $(-4,0)$
