The length of the latus rectum of an ellipse | vedclass

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(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The length of the latus rectum is $\frac{2b^2}{a}$ and th

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$\sqrt{\frac{2}{3}}$

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(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The length of the latus rectum is $\frac{2b^2}{a}$ and the length of the major axis is $2a$.Given that the length of the latus rectum is $\frac{1}{3}$ of the major axis:$\frac{2b^2}{a} = \frac{1}{3} (2a)$$\frac{b^2}{a} = \frac{a}{3}$$3b^2 = a^2$We know that for an ellipse,$b^2 = a^2(1 - e^2)$,where $e$ is the eccentricity.Substituting $b^2$ in the equation:$3a^2(1 - e^2) = a^2$$3(1 - e^2) = 1$$1 - e^2 = \frac{1}{3}$$e^2 = 1 - \frac{1}{3} = \frac{2}{3}$$e = \sqrt{\frac{2}{3}}$.

The length of the latus rectum of an ellipse is $\frac{1}{3}$ of its major axis. Its eccentricity is:

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