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(A) The vertices of the ellipse are given as $(\pm a, 0) = (\pm 5, 0)$,so $a = 5$.The foci are given as $(\pm ae, 0) = (\pm 4, 0)$,so $ae = 4$.Substit

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$9x^2 + 25y^2 = 225$

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(A) The vertices of the ellipse are given as $(\pm a, 0) = (\pm 5, 0)$,so $a = 5$.The foci are given as $(\pm ae, 0) = (\pm 4, 0)$,so $ae = 4$.Substituting $a = 5$,we get $5e = 4$,which means $e = \frac{4}{5}$.For an ellipse,the relationship between $a, b,$ and $e$ is $b^2 = a^2(1 - e^2)$.$b^2 = 5^2(1 - (\frac{4}{5})^2) = 25(1 - \frac{16}{25}) = 25(\frac{9}{25}) = 9$.Thus,$b = 3$.The standard equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Substituting the values,we get $\frac{x^2}{25} + \frac{y^2}{9} = 1$.Multiplying by $225$,we obtain $9x^2 + 25y^2 = 225$.

The equation of the ellipse whose vertices are $(\pm 5, 0)$ and foci are $(\pm 4, 0)$ is

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