The ellipse having its foci (0, ± 1) and major axis of length 5 is

Since the foci of given ellipse lie on the $Y$-axis, it is vertical ellipse. Let the required equation be $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ Where $a^2>b^2$ Foci are $(0, \pm c)=(0, \pm 1) \Rightarrow c=1a=$ length of the semi-major axis $=\frac{1}{2} \times \sqrt{5}=\frac{\sqrt{5}}{2}\therefore \quad a^2=\frac{5}{4}$ Now, $\quad c^2=a^2-b^2 \Rightarrow 1=\left(\frac{\sqrt{5}}{2}\right)^2-b^2\Rightarrow \quad b^2=\frac{5}{4}-1=\frac{1}{4}\therefore$ The required equation is $\frac{x^2}{1 / 4}+\frac{y^2}{5 / 4}=1 \Rightarrow 4 x^2+\frac{4 y^2}{5}=1\Rightarrow \quad 20 x^2+4 y^2=5$