If the eccentricity and the length of the lat | vedclass

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(A) Given,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. Eccentricity $e = \frac{\sqrt{3}}{2}$ and length of la

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$6$

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(A) Given,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. Eccentricity $e = \frac{\sqrt{3}}{2}$ and length of latus rectum $L = \frac{2b^2}{a} = 1$. We know that $b^2 = a^2(1 - e^2)$. Substituting $e^2 = \frac{3}{4}$,we get $b^2 = a^2(1 - \frac{3}{4}) = \frac{a^2}{4}$,which implies $\frac{b^2}{a^2} = \frac{1}{4}$ or $b^2 = \frac{a^2}{4}$. Substitute $b^2 = \frac{a^2}{4}$ into the latus rectum equation: $\frac{2(a^2/4)}{a} = 1$. This simplifies to $\frac{a}{2} = 1$,so $a = 2$. Then $b^2 = \frac{2^2}{4} = 1$,so $b = 1$. The length of the major axis is $2a = 2(2) = 4$. The length of the minor axis is $2b = 2(1) = 2$. The sum of the lengths of the major and minor axes is $4 + 2 = 6$.

If the eccentricity and the length of the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are $\frac{\sqrt{3}}{2}$ and $1$ respectively,then the sum of the lengths of the major axis and minor axis of the ellipse is

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