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

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(D) 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 $2b$.The distance between

Vedclass · Accepted Answer

$\frac{2}{\sqrt{5}}$

Vedclass · Answer

(D) 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 $2b$.The distance between the foci is $2ae$.According to the problem,$2b = \frac{1}{2} (2ae) = ae$.Thus,$\frac{b}{a} = \frac{e}{2}$.We know that for an ellipse,$b^2 = a^2(1 - e^2)$,which implies $\frac{b^2}{a^2} = 1 - e^2$.Substituting $\frac{b}{a} = \frac{e}{2}$,we get $(\frac{e}{2})^2 = 1 - e^2$.$\frac{e^2}{4} = 1 - e^2$.$e^2 + \frac{e^2}{4} = 1$.$\frac{5e^2}{4} = 1$.$e^2 = \frac{4}{5}$.$e = \frac{2}{\sqrt{5}}$.

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

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