Answer
Center at $\quad(0,0).$
Vertices at $\quad(\pm 2\sqrt{2},0).$
Foci at $\quad(\pm\sqrt{6},0)$
Work Step by Step
Divide the equation with $8$ ... (we want the standard form)
$\displaystyle \frac{4y^{2}}{8}+\frac{x^{2}}{8}=\frac{8}{8}$
$\displaystyle \frac{x^{2}}{8}+\frac{y^{2}}{2}=1$
$ a=\sqrt{8}=2\sqrt{2}\approx 2.828,\quad$ main axis: x-axis.
$b=\sqrt{2}$ .
Table 3: Major axis horizontal (parallel to the x-axis)$\begin{array}{lll}
\text{ Foci}&\text{ Vertices}&\text{ Equation}\\
{(h+c, k)}&{(h+a,k)}&{\displaystyle \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1}\\
{(h-c,k)}&{(h-a,k)}&{a \gt b \gt 0\text{ and }b^{2}=a^{2}-c^{2}}\\ \end{array}$
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The center of the ellipse is at the origin.
The vertices are $(2\sqrt{2},0)$ and $(-2\sqrt{2},0).$
Find $c:$
$c^{2}=a^{2}-b^{2}$
$c^{2}=8-2$
$c^{2}=6$
$c=\sqrt{6}\approx 2.449$
The foci are $(\sqrt{6},0)$ and $(-\sqrt{6},0)$
