Answer
$ \quad \displaystyle \frac{x^{2}}{3}+\frac{y^{2}}{4}=1$
Work Step by Step
The foci, center and vertices lie on the same line, the main axis.
Main axis here: $x=0$ (the $y$-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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Center: $(0,0) \quad $
Focus: $(0,1) \quad \Rightarrow \quad c=1$
Vertex: $(0,-2) \quad \Rightarrow \quad a=2$
Find b:
$b^{2}=a^{2}-c^{2}=4-1=3$
$b=\sqrt{3} \quad \approx 1.732$
The equation is $ \quad \displaystyle \frac{x^{2}}{3}+\frac{y^{2}}{4}=1$
