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

#### Solution

\[\text{ According to the Pythogoras theorem, we have }:\]

\[O A^2 + O B^2 = A B^2 \]

\[\text{ From the figure, we can see that } \]

\[OA = \sqrt{\left( \frac{b^2}{a} - 0 \right)^2 + \left( ae - 0 \right)^2} = \sqrt{\frac{b^4}{a^2}} + a^2 e^2 = OB and AB = \frac{2 b^2}{a}\]

\[\text{ Now }, 2\left[ a^2 e^2 + \frac{b^4}{a^2} \right] = \frac{4 b^4}{a^2}\]

\[ \Rightarrow a^2 e^2 + \frac{b^4}{a^2} = \frac{2 b^4}{a^2}\]

\[ \Rightarrow a^2 e^2 = - \frac{b^4}{a^2} + \frac{2 b^4}{a^2}\]

\[ \Rightarrow a^2 e^2 = \frac{b^4}{a^2}\]

\[ \Rightarrow e^2 = \frac{b^4}{a^4}\]

\[ \Rightarrow e = \frac{b^2}{a^2}\]

\[\text{ We know that } e = \sqrt{1 - \frac{b^2}{a^2}}\]

\[e = \sqrt{1 - e}\]

\[\text{ On squaring both sides, we get }:\]

\[ e^2 + e - 1 = 0\]

\[ \Rightarrow e = \frac{- 1 \pm \sqrt{1 + 4}}{2} \left( \because \text{ Ecentricity cannot be negative } \right)\]

\[ \Rightarrow e = \frac{\sqrt{5} - 1}{2} \]