The equation of ellipse whose focus is $\left(0, \sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\right),$ directrix is $\mathrm{y}=\frac{{\mathrm{a}}^2}{\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}}$ and eccentricity is $\frac{\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}}{\mathrm{a}}$ is

$$\begin{gathered} Let the ellipse be \frac{{\mathrm{x}}^2}{{\mathrm{A}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{B}}^2}=1(A<B) \\ Given \\ \mathrm{focus}=(0, Be)=(\mathrm{o}, \sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}) \\ \Rightarrow Be=\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2} \\ \Rightarrow \mathrm{B}\left(\frac{\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}}{\mathrm{a}}\right)=\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2} \\ \Rightarrow \mathrm{B}=\mathrm{a} \\ Now {\mathrm{A}}^2={\mathrm{B}}^2(1-{\mathrm{e}}^2) \\ ={\mathrm{a}}^2\left(1-\frac{({\mathrm{a}}^2-{\mathrm{b}}^2)}{{\mathrm{a}}^2}\right)={\mathrm{b}}^2 \\ \therefore \mathrm{eq}uation \mathrm{of} \mathrm{ellipse} is \frac{{\mathrm{x}}^2}{{\mathrm{A}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{B}}^2}=1 \\ \Rightarrow \frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}=1 \end{gathered}$$