The foci of the ellipse 25(x + 1)^2 + 9(y + 2 | vedclass

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(A) The given equation is $25(x + 1)^2 + 9(y + 2)^2 = 225$.Dividing by $225$,we get $\frac{(x + 1)^2}{9} + \frac{(y + 2)^2}{25} = 1$.Here,$a^2 = 25$ a

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

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(A) The given equation is $25(x + 1)^2 + 9(y + 2)^2 = 225$.Dividing by $225$,we get $\frac{(x + 1)^2}{9} + \frac{(y + 2)^2}{25} = 1$.Here,$a^2 = 25$ and $b^2 = 9$,so $a = 5$ and $b = 3$.Since $a > b$,the major axis 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 of the ellipse is $(h, k) = (-1, -2)$.The foci are given by $(h, k \pm ae) = (-1, -2 \pm 5 	imes \frac{4}{5}) = (-1, -2 \pm 4)$.Thus,the foci are $(-1, -2 + 4) = (-1, 2)$ and $(-1, -2 - 4) = (-1, -6)$.

The foci of the ellipse $25(x + 1)^2 + 9(y + 2)^2 = 225$ are at

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