Answer
Center at $\quad(0,0).$
Vertices at $\quad(0,\pm 3).$
Foci at $\quad(0,\pm\sqrt{5})$
Work Step by Step
Divide the equation with $36$ ... (we want the standard form)
$\displaystyle \frac{4y^{2}}{36}+\frac{9x^{2}}{36}=1$
$\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{9}=1$
$\displaystyle \frac{x^{2}}{2^{2}}+\frac{y^{2}}{3^{2}}=1,\quad$ main axis: $y$-axis$, a=3,\ b=2.$
Table 3: Major axis vertical (parallel to the y-axis)
$\begin{array}{lll}
\text{ Foci}&\text{ Vertices}&\text{ Equation}\\
{(h,k+c)}&{(h,k+a)}&{\displaystyle \frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1}\\
{(h,k-c)}&{(h,k-a)}&{a \gt b \gt 0\text{ and }b^{2}=a^{2}-c^{2}}\end{array}$
The center of the ellipse is at the origin.
The vertices are $(0,3)$ and $(0,-3).$
Find $c:$
$c^{2}=a^{2}-b^{2}$
$c^{2}=9-4$
$c^{2}=5\quad $
$c=\sqrt{5}\approx 2.236$
The foci are $(0,\sqrt{5})$ and $(0,-\sqrt{5})$