Q

# 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)

$1/9$

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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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