Question

The eccentricity of the ellipse, if the distance between the foci is equal to the length of the latus-rectum, is

Options

• \[\frac{\sqrt{5} - 1}{2}\]

   

• \[\frac{\sqrt{5} + 1}{2}\]

   

• \[\frac{\sqrt{5} - 1}{4}\]

   

• none of these

   

Answer 1

\[e = \frac{\sqrt{5} - 1}{2} \]
According to the question, the distance between the foci is equal to the length of the latus rectum.
\[\frac{2 b^2}{a} = 2ae\]
\[ \Rightarrow b^2 = a^2 e\]
\[\text{ Now, }e = \sqrt{1 - \frac{b^2}{a^2}}\]
\[ \Rightarrow e = \sqrt{1 - \frac{a^2 e}{a^2}}\]
\[ \Rightarrow e = \sqrt{1 - e}\]
On squaring both sides, we get:
\[ e^2 + e - 1 = 0\]
\[ \Rightarrow e = \frac{- 1 \pm \sqrt{1 + 4}}{2}\]
\[ \Rightarrow e = \frac{\sqrt{5} - 1}{2} \left( \because\text{ e cannot be negative }\right)\]