The foci of the ellipse 9x^2 + 25y^2 = 225 ar | vedclass

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(A) The given equation of the ellipse is $9x^2 + 25y^2 = 225$. Dividing both sides by $225$,we get $\frac{9x^2}{225} + \frac{25y^2}{225} = 1$,which si

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$(\pm 4, 0)$

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(A) The given equation of the ellipse is $9x^2 + 25y^2 = 225$. Dividing both sides by $225$,we get $\frac{9x^2}{225} + \frac{25y^2}{225} = 1$,which simplifies to $\frac{x^2}{25} + \frac{y^2}{9} = 1$. Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 25$ and $b^2 = 9$,so $a = 5$ and $b = 3$. The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$. The coordinates of the foci are $(\pm ae, 0)$. Substituting the values,we get $(\pm 5 	imes \frac{4}{5}, 0) = (\pm 4, 0)$.

The foci of the ellipse $9x^2 + 25y^2 = 225$ are

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