Question

If a latus rectum of an ellipse subtends a right angle at the centre of the ellipse, then write the eccentricity of the ellipse.

Answer 1

$$\begin{gathered} \text{According to the Pythogoras theorem, we have:} \\ O{A}^2+O{B}^2=A{B}^2 \\ \mathrm{From}\mathrm{the}\mathrm{figure},\mathrm{we}\mathrm{can}\mathrm{see}\mathrm{that} \\ OA=\sqrt{{\left(\frac{b^2}{a}-0\right)}^2+{\left(ae-0\right)}^2}=\sqrt{\frac{b^4}{a^2}}+a^2{e}^2=\mathrm{OB}\mathrm{and}\mathrm{AB}=\frac{2b^2}{a} \\ \mathrm{Now},2\left[a^2{e}^2+\frac{b^4}{a^2}\right]=\frac{4b^4}{a^2} \\ \Rightarrow a^2{e}^2+\frac{b^4}{a^2}=\frac{2b^4}{a^2} \\ \Rightarrow a^2{e}^2=-\frac{b^4}{a^2}+\frac{2b^4}{a^2} \\ \Rightarrow a^2{e}^2=\frac{b^4}{a^2} \\ \Rightarrow e^2=\frac{b^4}{a^4} \\ \Rightarrow e=\frac{b^2}{a^2} \\ \mathrm{We}\mathrm{know}\mathrm{that}e=\sqrt{1-\frac{b^2}{a^2}} \\ e=\sqrt{1-e} \\ \text{On squaring both sides, we get:} \\ {e}^2+e-1=0 \\ \Rightarrow e=\frac{-1\pm \sqrt{1+4}}{2}\left(\because \mathrm{Ecentricity}\mathrm{cannot}\mathrm{be}\mathrm{negative}\right) \\ \Rightarrow e=\frac{\sqrt{5}-1}{2} \end{gathered}$$