If the length of the latus rectum of an ellip | vedclass

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(D) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The distance between a focus $(ae, 0)$ and its nearest vertex $(a, 0)$ is

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(D) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The distance between a focus $(ae, 0)$ and its nearest vertex $(a, 0)$ is $a(1 - e) = \frac{3}{2}$.Thus,$a - ae = \frac{3}{2}$,which implies $ae = a - \frac{3}{2}$.The length of the latus rectum is $\frac{2b^2}{a} = 4$,so $b^2 = 2a$.Using the relation $b^2 = a^2(1 - e^2)$,we have $2a = a^2(1 - e^2)$,which simplifies to $1 - e^2 = \frac{2}{a}$,or $e^2 = 1 - \frac{2}{a}$.From $ae = a - \frac{3}{2}$,squaring both sides gives $a^2e^2 = (a - \frac{3}{2})^2$.Substituting $e^2 = 1 - \frac{2}{a}$,we get $a^2(1 - \frac{2}{a}) = a^2 - 3a + \frac{9}{4}$.$a^2 - 2a = a^2 - 3a + \frac{9}{4}$.$a = \frac{9}{4}$.Now,$e^2 = 1 - \frac{2}{a} = 1 - \frac{2}{9/4} = 1 - \frac{8}{9} = \frac{1}{9}$.Therefore,$e = \frac{1}{3}$.

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 $\frac{3}{2} \ units$,then its eccentricity is?

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