If the length of the minor axis of an ellipse | vedclass

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(A) 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 minor axis is $2b$ and the distance bet

Vedclass · Accepted Answer

$\frac{4}{\sqrt{17}}$

Vedclass · Answer

(A) 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 minor axis is $2b$ and the distance between the foci is $2ae$.Given that $2b = \frac{1}{4}(2ae)$,we have $b = \frac{ae}{4}$,which implies $\frac{b}{a} = \frac{e}{4}$.We know that for an ellipse,$e^2 = 1 - \frac{b^2}{a^2}$.Substituting $\frac{b}{a} = \frac{e}{4}$,we get $e^2 = 1 - \left(\frac{e}{4}\right)^2$.$e^2 = 1 - \frac{e^2}{16}$.$e^2 + \frac{e^2}{16} = 1$.$\frac{17e^2}{16} = 1$.$e^2 = \frac{16}{17}$.Thus,$e = \frac{4}{\sqrt{17}}$.

If the length of the minor axis of an ellipse is equal to one-fourth of the distance between the foci,then the eccentricity of the ellipse is:

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