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Need clarity, kindly explain! If the length of the latus rectum of an ellipse is 4 units and the distance between a focusand its nearest vertex on the major axis is3/2units, then its eccentricity is :

If the length of the latus rectum of an ellipse is 4 units and the distance between a focus
and its nearest vertex on the major axis is
3/2
units, then its eccentricity is :

  • Option 1)

    1/2

  • Option 2)

    1/3

  • Option 3)

    2/3

  • Option 4)

    1/9

 
Answers (2)
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As we have learned

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

 

 

Sum of focal distance -

2a

- wherein

 

 

Length of LR \frac{2b^{2}}{a}= 4 \Rightarrow b^{2}= 2a

also a- ae = 3/2\Rightarrow a(1-e)=3/2

also b^{2}= a^{2}(1-e^{2})= 2a

a(1-e)(1+e) = 2

thus a(1-e)= \frac{2}{1+e}

thus \frac{2}{1+e}= 3/2\Rightarrow e= 1/3

 

 


Option 1)

1/2

This is incorrect

Option 2)

1/3

This is correct

Option 3)

2/3

This is incorrect

Option 4)

1/9

This is incorrect

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