The foci of the ellipse 2x^2 + 3y^2 - 4x - 12 | vedclass

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(A) Given the equation of the ellipse: $2x^2 + 3y^2 - 4x - 12y + 13 = 0$.Rearranging the terms to complete the square:$2(x^2 - 2x) + 3(y^2 - 4y) = -13

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$\left(1 + \frac{1}{\sqrt{6}}, 2\right)$ and $\left(1 - \frac{1}{\sqrt{6}}, 2\right)$

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(A) Given the equation of the ellipse: $2x^2 + 3y^2 - 4x - 12y + 13 = 0$.Rearranging the terms to complete the square:$2(x^2 - 2x) + 3(y^2 - 4y) = -13$$2(x^2 - 2x + 1) + 3(y^2 - 4y + 4) = -13 + 2(1) + 3(4)$$2(x - 1)^2 + 3(y - 2)^2 = 1$Dividing by $1$,we get the standard form:$\frac{(x - 1)^2}{1/2} + \frac{(y - 2)^2}{1/3} = 1$Here,$a^2 = 1/2$ and $b^2 = 1/3$. Since $a^2 > b^2$,the major axis is horizontal.The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1/3}{1/2}} = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$.The distance from the center to the foci is $ae = \sqrt{1/2} 	imes \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{6}}$.The center is $(h, k) = (1, 2)$.The foci are $(h \pm ae, k) = \left(1 \pm \frac{1}{\sqrt{6}}, 2ight)$.

The foci of the ellipse $2x^2 + 3y^2 - 4x - 12y + 13 = 0$ are

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