The major and minor axes of an ellipse are al | vedclass

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(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. The length of the latus rectum is $\frac{2b^2}{a} = 4$,w

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

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(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. 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,$a^2e^2 = 8$. Since $e^2 = 1 - \frac{b^2}{a^2}$,we have $a^2(1 - \frac{b^2}{a^2}) = 8$,which simplifies to $a^2 - b^2 = 8$. Substituting $b^2 = 2a$,we get $a^2 - 2a - 8 = 0$. Factoring the quadratic equation,$(a - 4)(a + 2) = 0$. Since $a > 0$,we have $a = 4$. Then $b^2 = 2(4) = 8$. The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{8} = 1$,which simplifies to $x^2 + 2y^2 = 16$.

The major and minor axes of an ellipse are along the $X$-axis and $Y$-axis respectively. If its latus rectum is of length $4$ and the distance between the foci is $4 \sqrt{2}$,then the equation of that ellipse is

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