The foci of the conic 25x^2 + 16y^2 - 150x = | vedclass

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(B) Given equation: $25x^2 + 16y^2 - 150x = 175$. Completing the square for $x$: $25(x^2 - 6x) + 16y^2 = 175$. $25(x^2 - 6x + 9) + 16y^2 = 175 + 225$.

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$(3, 3)$ and $(3, -3)$

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(B) Given equation: $25x^2 + 16y^2 - 150x = 175$. Completing the square for $x$: $25(x^2 - 6x) + 16y^2 = 175$. $25(x^2 - 6x + 9) + 16y^2 = 175 + 225$. $25(x - 3)^2 + 16y^2 = 400$. Dividing by $400$: $\frac{(x - 3)^2}{16} + \frac{y^2}{25} = 1$. This is an ellipse with center $(h, k) = (3, 0)$,$a^2 = 25$,and $b^2 = 16$. Since $a^2 > b^2$,the major axis is vertical. Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$. Foci are $(h, k \pm ae) = (3, 0 \pm 5 	imes \frac{3}{5}) = (3, \pm 3)$.

The foci of the conic $25x^2 + 16y^2 - 150x = 175$ are

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