In an ellipse,its foci and the ends of its ma | vedclass

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(D) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a > b$.The ends of the major axis are $A'(-a, 0)$ and $A(a, 0)$.The foci are $S'

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$3$

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(D) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a > b$.The ends of the major axis are $A'(-a, 0)$ and $A(a, 0)$.The foci are $S'(-ae, 0)$ and $S(ae, 0)$.Given that the foci and the ends of the major axis are equally spaced,we have the segments $A'S'$,$S'S$,and $SA$ equal in length.$A'S' = S'S = SA = k$ (say).The total length of the major axis is $A'A = 2a$.Thus,$k + k + k = 2a \implies 3k = 2a \implies k = \frac{2a}{3}$.The distance between the foci is $S'S = 2ae = k = \frac{2a}{3}$.Therefore,$e = \frac{1}{3}$.We know the relation $b^2 = a^2(1 - e^2)$.Given $b = 2\sqrt{2}$,so $b^2 = 8$.$8 = a^2(1 - (\frac{1}{3})^2) = a^2(1 - \frac{1}{9}) = a^2(\frac{8}{9})$.$8 = \frac{8a^2}{9} \implies a^2 = 9 \implies a = 3$.The length of the semi-major axis is $3$.

In an ellipse,its foci and the ends of its major axis are equally spaced. If the length of its semi-minor axis is $2 \sqrt{2}$,then the length of its semi-major axis is

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