Answer : $\large\frac{4x^2}{1}+\frac{4y^2}{5}\normalsize =1$
Given foci is $(0,\pm 1)$
Clearly the ellipse lies along the minor axis.
Length of the minor axis is $\sqrt 5$
(ie) $2b=\sqrt 5$
$\therefore b=\large\frac{\sqrt 5}{2}$
Given foci is $be=1$
$\Rightarrow \large\frac{\sqrt 5}{2}$$\times e=1$
$\therefore e=\large\frac{2}{\sqrt 5}$
Hence $a^2=b^2(1-e^2)$
$a^2=\large\frac{5}{4}$$(1-\large\frac{4}{5})$
$a^2=\large\frac{1}{4}$ and $b^2=\large\frac{5}{4}$
Hence equation of the ellipse is $\large\frac{x^2}{\Large\frac{1}{4}}+\frac{y^2}{\Large\frac{5}{4}}$$=1$
$\Rightarrow \large\frac{4x^2}{1}+\frac{4y^2}{5}$$=1$ is the required equation.