Answer
$ \quad \displaystyle \frac{x^{2}}{4}+(y-1)^{2}=1$
Work Step by Step
1.$ \quad $Major axis: parallel to $x$ -axis$ \quad \Rightarrow \quad $see table 3
$ \displaystyle \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1, \quad a \gt b \gt 0$
2.$ \quad $ Center: $(0,1)$
3.$ \quad 2a=4 \quad $ (Length of major axis)$ \quad a=2$
4.$ \quad 2b=1 \quad $ (Length of minor axis)$ \quad b=1$
Equation:$ \quad \displaystyle \frac{(x-0)^{2}}{4}+\frac{(y-1)^{2}}{1}=1$
or $ \quad \displaystyle \frac{x^{2}}{4}+(y-1)^{2}=1$
