The coordinates of the foci of the ellipse 3x | vedclass

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(C) Given equation: $3x^2 + 4y^2 - 12x - 8y + 4 = 0$Rearranging terms: $3(x^2 - 4x) + 4(y^2 - 2y) = -4$Completing the square: $3(x - 2)^2 - 12 + 4(y -

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$(1, 1), (3, 1)$

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(C) Given equation: $3x^2 + 4y^2 - 12x - 8y + 4 = 0$Rearranging terms: $3(x^2 - 4x) + 4(y^2 - 2y) = -4$Completing the square: $3(x - 2)^2 - 12 + 4(y - 1)^2 - 4 = -4$$3(x - 2)^2 + 4(y - 1)^2 = 12$Dividing by $12$: $\frac{(x - 2)^2}{4} + \frac{(y - 1)^2}{3} = 1$Here,$a^2 = 4$ and $b^2 = 3$,so $a = 2$ and $b = \sqrt{3}$.The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.The foci in the shifted coordinate system $(X, Y)$ are $(\pm ae, 0)$,where $X = x - 2$ and $Y = y - 1$.$ae = 2 	imes \frac{1}{2} = 1$.So,$X = \pm 1$ and $Y = 0$.$x - 2 = 1 \implies x = 3$ and $y - 1 = 0 \implies y = 1$.$x - 2 = -1 \implies x = 1$ and $y - 1 = 0 \implies y = 1$.The foci are $(3, 1)$ and $(1, 1)$.

The coordinates of the foci of the ellipse $3x^2 + 4y^2 - 12x - 8y + 4 = 0$ are

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