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(C) Given,length of latus rectum = $\frac{2b^2}{a} = 4$,so $b^2 = 2a$. Distance between foci = $2ae = 4\sqrt{2}$,so $ae = 2\sqrt{2}$. We know $b^2 = a

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$x^2 + 2y^2 = 16$

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(C) Given,length of latus rectum = $\frac{2b^2}{a} = 4$,so $b^2 = 2a$. Distance between foci = $2ae = 4\sqrt{2}$,so $ae = 2\sqrt{2}$. We know $b^2 = a^2(1 - e^2) = a^2 - a^2e^2$. Substituting $b^2 = 2a$ and $ae = 2\sqrt{2}$ (so $a^2e^2 = 8$): $2a = a^2 - 8$ $a^2 - 2a - 8 = 0$ $(a - 4)(a + 2) = 0$. Since $a > 0$,we have $a = 4$. Then $b^2 = 2(4) = 8$. The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{16} + \frac{y^2}{8} = 1$. Multiplying by $16$,we get $x^2 + 2y^2 = 16$.

The equation of the ellipse referred to its axes as the axes of coordinates with latus rectum of length $4$ and distance between foci $4 \sqrt{2}$ is-

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