The latus rectum of an ellipse is 10 and the | vedclass

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(A) Given the length of the latus rectum is $\frac{2b^2}{a} = 10$. Given the minor axis is equal to the distance between the foci,so $2b = 2ae$,which

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$x^2 + 2y^2 = 100$

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(A) Given the length of the latus rectum is $\frac{2b^2}{a} = 10$. Given the minor axis is equal to the distance between the foci,so $2b = 2ae$,which implies $b = ae$. We know that $b^2 = a^2(1 - e^2)$. Substituting $b = ae$,we get $a^2e^2 = a^2(1 - e^2)$,which simplifies to $e^2 = 1 - e^2$,so $2e^2 = 1$ or $e = \frac{1}{\sqrt{2}}$. Since $b = ae$,we have $b^2 = a^2e^2 = a^2(\frac{1}{2})$. Substituting this into the latus rectum equation: $\frac{2(a^2/2)}{a} = 10$,which gives $a = 10$. Then $b^2 = \frac{100}{2} = 50$,so $b = \sqrt{50} = 5\sqrt{2}$. The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{100} + \frac{y^2}{50} = 1$. Multiplying by $100$,we get $x^2 + 2y^2 = 100$.

The latus rectum of an ellipse is $10$ and the minor axis is equal to the distance between the foci. The equation of the ellipse is

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