If distance between foci of an ellipse is half length of its latus rectum, then eccentricity of elli

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10-2. Parabola, Ellipse, Hyperbola

hard

If the distance between the foci of an ellipse is half the length of its latus rectum, then the eccentricity of the ellipse is

A

$\frac{{2\sqrt 2 - 1}}{2}$

B

$\sqrt 2 - 1$

C

$\frac{1}{2}$

D

$\frac{{\sqrt 2 - 1}}{2}$

(JEE MAIN-2015)

Solution

Focus of an ellipse is given as $\left( { \pm ae,0} \right)$

distance between then $=2ae$

According to the quation, $2ae = \frac{{{b^2}}}{a}$

$ \Rightarrow 2{a^2}e = {b^2} = {a^2}\left( {1 – {e^2}} \right)$

$ \Rightarrow 2e = 1 – {e^2} \Rightarrow {\left( {e – 1} \right)^2} = 2 \Rightarrow e = \sqrt 2 – 1$

Standard 11

Mathematics

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