Question

find eccentricity of an ellipse if its latus rectum is 1/3rd of its minor axis

Answer 1

Dear student
$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{ellipse}\mathrm{be}\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ \mathrm{Its}\mathrm{latus}\mathrm{rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\mathrm{and}\mathrm{its}\mathrm{minor}\mathrm{axis}=2\mathrm{b} \\ \mathrm{Given}\mathrm{latus}\mathrm{rectum}=\left(\frac{1}{3}\right)\left(\mathrm{minor}\mathrm{axis}\right) \\ \mathrm{So},\frac{2{\mathrm{b}}^2}{\mathrm{a}}=\frac{1}{3}\left(2\mathrm{b}\right) \\ \Rightarrow \mathrm{a}=3\mathrm{b} \\ \mathrm{Thus}{\mathrm{e}}^2=\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{9{\mathrm{b}}^2-{\mathrm{b}}^2}{9{\mathrm{b}}^2}=\frac{8}{9} \\ \mathrm{So},\mathrm{e}=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3} \end{gathered}$$
Regards