The foci of 16x^2 + 25y^2 = 400 are | vedclass

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(A) The given equation of the ellipse is $16x^2 + 25y^2 = 400$.Dividing both sides by $400$,we get $\frac{x^2}{25} + \frac{y^2}{16} = 1$.Comparing thi

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$(\pm 3, 0)$

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(A) The given equation of the ellipse is $16x^2 + 25y^2 = 400$.Dividing both sides by $400$,we get $\frac{x^2}{25} + \frac{y^2}{16} = 1$.Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 25$ and $b^2 = 16$,so $a = 5$ and $b = 4$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.The foci are given by $(\pm ae, 0) = (\pm 5 	imes \frac{3}{5}, 0) = (\pm 3, 0)$.

The foci of $16x^2 + 25y^2 = 400$ are

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