Foci of the ellipse $3 x^{2} + 4 y^{2} - 12 x - 8 y + 4 = 0$ are:

(2, 1), (4, 1)

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(1, 1), (3, 1)

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(3, 1), (5, 1)

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(- 4, 1), (5, 1)

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Solution

The correct option is B $(1, 1), (3, 1)$
We have,
$3 x^{2} + 4 y^{2} - 12 x - 8 y + 4 = 0$

$\Rightarrow 3 (x^{2} - 4 x) + 4 (y^{2} - 2 y) = - 4$

$\Rightarrow 3 (x^{2} - 4 x + 4) + 4 (y^{2} - 2 y + 1) = - 4 + 12 + 4$

$\Rightarrow 3 (x - 2)^{2} + 4 (y - 1)^{2} = 12$

$\Rightarrow \frac{(x - 2)^{2}}{4} + \frac{(y - 1)^{2}}{3} = 1$

$\Rightarrow a^{2} = 4, b^{2} = 3 ∴ e = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$

Hence foci are $(x - 2, y - 1) = (\pm a e, 0) = (\pm 1, 0)$

$\Rightarrow (x, y) = (2 \pm 1, 1) :: (3, 1) o r (1, 1)$

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