An ellipse passing through (4 sqrt{2}, 2 sqrt | vedclass

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(B) The $y$-coordinate of the foci is $0$,so the major axis is on the $X$-axis. \ Given $ae = 4$. \ Let the equation of the ellipse be $\frac{x^2}{a

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$\frac{1}{2}$

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(B) The $y$-coordinate of the foci is $0$,so the major axis is on the $X$-axis. \ Given $ae = 4$. \ Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. \ Since $b^2 = a^2(1 - e^2) = a^2 - (ae)^2 = a^2 - 16$,we substitute the point $(4\sqrt{2}, 2\sqrt{6})$: \ $\frac{(4\sqrt{2})^2}{a^2} + \frac{(2\sqrt{6})^2}{a^2 - 16} = 1$ \ $\frac{32}{a^2} + \frac{24}{a^2 - 16} = 1$ \ $32(a^2 - 16) + 24a^2 = a^2(a^2 - 16)$ \ $32a^2 - 512 + 24a^2 = a^4 - 16a^2$ \ $a^4 - 72a^2 + 512 = 0$ \ $(a^2 - 64)(a^2 - 8) = 0$ \ Since $a > ae$,$a^2$ must be greater than $16$,so $a^2 = 64$ and $a = 8$. \ Using $ae = 4$,we get $8e = 4$,which implies $e = \frac{1}{2}$.

An ellipse passing through $(4 \sqrt{2}, 2 \sqrt{6})$ has foci at $(-4, 0)$ and $(4, 0)$. Then,its eccentricity is

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