Find the coordinates of the foci of the ellip | vedclass

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(A) Given the equation of the ellipse: $25(x + 1)^2 + 9(y + 2)^2 = 225$.Divide by $225$ to get the standard form:$\frac{(x + 1)^2}{9} + \frac{(y + 2)^

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$(-1, 2)$ and $(-1, -6)$

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(A) Given the equation of the ellipse: $25(x + 1)^2 + 9(y + 2)^2 = 225$.Divide by $225$ to get the standard form:$\frac{(x + 1)^2}{9} + \frac{(y + 2)^2}{25} = 1$.Here,$a^2 = 25$ and $b^2 = 9$,so $a = 5$ and $b = 3$.Since $a > b$,the ellipse is vertical.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 center is $(h, k) = (-1, -2)$.The foci are $(h, k \pm ae) = (-1, -2 \pm 5 	imes \frac{4}{5}) = (-1, -2 \pm 4)$.Thus,the foci are $(-1, 2)$ and $(-1, -6)$.

Find the coordinates of the foci of the ellipse $25(x + 1)^2 + 9(y + 2)^2 = 225$.

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