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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 distance between the two foci is $2ae$.The length of

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$e = \frac{1}{\sqrt{2}}$

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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 distance between the two foci is $2ae$.The length of the minor axis is $2b$.According to the problem,$2ae = 2b$,which implies $ae = b$.Squaring both sides,we get $a^2e^2 = b^2$.We know that for an ellipse,$b^2 = a^2(1 - e^2)$.Substituting $b^2$,we get $a^2e^2 = a^2(1 - e^2)$.Dividing by $a^2$,we get $e^2 = 1 - e^2$.$2e^2 = 1 \Rightarrow e^2 = \frac{1}{2}$.Therefore,$e = \frac{1}{\sqrt{2}}$.

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

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