If $x^{2} + 4 y^{2} - 8 x + 12 = 0$ is satisfied by real values of $x & y,$ then $y \in$

$[- 1, 1]$

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$[2, 6]$

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$[- 2, - 1]$

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$[2, 5]$

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Solution

The correct option is A
$[- 1, 1]$

Given : $x^{2} + 4 y^{2} - 8 x + 12 = 0$
$\Rightarrow x^{2} - 8 x + 4 (y^{2} + 3)$

Also given that it only satisfy for real values.
$∴ D \geq 0$
$\Rightarrow (8)^{2} - 4 \cdot 4 (y^{2} + 3) \geq 0$
$\Rightarrow 64 - 16 y^{2} - 48 \geq 0$
$\Rightarrow - 16 y^{2} + 16 \geq 0$
$\Rightarrow 16 y^{2} - 16 \leq 0$
$\Rightarrow y^{2} - 1 \leq 0$
$\Rightarrow (y + 1) (y - 1) \leq 0$
$\Rightarrow - 1 \leq y \leq 1$
Hence, $y \in [- 1, 1]$

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