The equation of the ellipse in the standard f | vedclass

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(C) The length of the latus rectum is $\frac{2b^2}{a} = 4$,which implies $b^2 = 2a$. The distance between the foci is $2ae = 4\sqrt{2}$,so $ae = 2\sqr

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

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(C) The length of the latus rectum is $\frac{2b^2}{a} = 4$,which implies $b^2 = 2a$. The distance between the foci is $2ae = 4\sqrt{2}$,so $ae = 2\sqrt{2}$. Squaring both sides,$a^2e^2 = 8$. Using the relation $a^2e^2 = a^2 - b^2$,we have $a^2 - b^2 = 8$. Substituting $b^2 = 2a$,we get $a^2 - 2a - 8 = 0$. Solving the quadratic equation $(a - 4)(a + 2) = 0$,we get $a = 4$ (since $a > 0$). 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 in the standard form whose length of the latus rectum is $4$ and whose distance between the foci is $4 \sqrt{2}$,is

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