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

Option 1)
$\frac{1}{2}\:$

Option 2)
$\frac{2\sqrt{2}-1}{2}\;$

Option 3)
$\sqrt{2}-1\;$

Option 4)
$\frac{\sqrt{2}-1}{2}$

Answers (1)

best_answer

As learnt in concept

Coordinates of foci -

$\pm ae,o$

- wherein

For the ellipse

$\frac{x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}= 1$

Length of latus rectum of ellipse -

$\frac{2b^{2}}{a}$

- wherein

For the ellipse

$\frac{x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}= 1$

Eccentricity -

$e= \sqrt{1-\frac{b^{2}}{a^{2}}}$

- wherein

For the ellipse

$\frac{x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}= 1$

Given that

$2ae=\frac{1}{2}\times \frac{2b^{2}}{a}$

$we \: get \:\:\frac{b^{2}}{a^{2}}=2e$

Also $e=\sqrt{1-\frac{b^{2}}{a^{2}}}$

$=> e=\sqrt{1-2e}$

Squaring both sides

$e^{2}+2e-1=0$

$e=\frac{-2^+_-\sqrt{4+4}}{2}$

$=-1^+_-\sqrt{2}$

$either \: e=\sqrt{2}-1$ $or\: e=\sqrt{2}+1$

for ellipse, $e=\sqrt{2}-1$

Option 1)

$\frac{1}{2}\:$

Incorrect option

Option 2)

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

Incorrect option

Option 3)

$\sqrt{2}-1\;$

Correct option

Option 4)

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

Incorrect option

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prateek

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