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)

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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