Answer
Center at $\quad(0,0).$
Vertices at $\quad(0,\pm 4).$
Foci at $\quad(0,\pm 2\sqrt{3})$
Work Step by Step
Divide the equation with 16 ... (we want the standard form)
$\displaystyle \frac{4x^{2}}{16}+\frac{y^{2}}{16}=1$
$\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{16}=1$
$\displaystyle \frac{x^{2}}{2^{2}}+\frac{y^{2}}{4^{2}}=1\qquad$... $a=4,\ b=2$, main axis: y-axis.
Table 3: Major axis vertical (parallel to the y-axis)
$\begin{array}{lll}
\text{ Foci}&\text{ Vertices}&\text{ Equation}\\
{(h,k+c)}&{(h,k+a)}&{\displaystyle \frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1}\\
{(h,k-c)}&{(h,k-a)}&{a \gt b \gt 0\text{ and }b^{2}=a^{2}-c^{2}}\end{array}$
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The center of the ellipse is at the origin.
The vertices are $\quad (0,4)$ and $(0,-4).$
Find $c:$
$c^{2}=a^{2}-b^{2}$
$c^{2}=16-4$
$c^{2}=12$
$c=\sqrt{12}=2\sqrt{3} \quad \approx 3.464$
The foci are $\quad (0,2\sqrt{3})$ and $(0,-2\sqrt{3}).$
