Let the length of the latus rectum of an elli | vedclass

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(C) Given the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$.Length of latus rectum $= \frac{2b^2}{a} = a \implies 2b^2 = a^2 \dots (i)$

Vedclass · Accepted Answer

$\frac{1}{e}$

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(C) Given the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$.Length of latus rectum $= \frac{2b^2}{a} = a \implies 2b^2 = a^2 \dots (i)$.The equation of the director circle is $x^2 + y^2 = a^2 + b^2$.Given the radius is $\sqrt{3}$,so $a^2 + b^2 = (\sqrt{3})^2 = 3$.Substituting $a^2 = 2b^2$ into the equation: $2b^2 + b^2 = 3 \implies 3b^2 = 3 \implies b^2 = 1$.Then $a^2 = 2(1) = 2$,so $a = \sqrt{2}$.The length of the latus rectum is $\frac{2b^2}{a} = \frac{2(1)}{\sqrt{2}} = \sqrt{2}$.Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}}$.Thus,$\frac{1}{e} = \sqrt{2}$.Therefore,the length of the latus rectum is $\frac{1}{e}$.

Let the length of the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be equal to the length of its semi-major axis. If the radius of its director circle is $\sqrt{3}$ and $e$ is its eccentricity,then the length of its latus rectum is

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