If $2 x^{2} y^{2} + y^{2} - 6 x^{2} - 12 = 0,$ then number of integral pairs $(x, y)$ satisfying is/are

1

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4

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Solution

The correct option is C $4$
$2 x^{2} y^{2} + y^{2} - 6 x^{2} - 12 = 0$
$\Rightarrow 2 x^{2} (y^{2} - 3) = 12 - y^{2}$
$\Rightarrow 2 x^{2} = \frac{12 - y^{2}}{y^{2} - 3}$
$\Rightarrow 2 x^{2} + 1 = \frac{9}{y^{2} - 3}$

Now, LHS is an odd positive number for integral values of $x .$

$∴ \frac{9}{y^{2} - 3}$ must be an odd positive integer for integral values of $y$

$∴ y^{2} = 4$ is the only possibility.
$\Rightarrow y = - 2, 2$
$∴ 2 x^{2} + 1 = 9$
$\Rightarrow x^{2} = 4$
$\Rightarrow x = 2, - 2$

$∴$ Possible pairs are $(2, 2), (2, - 2), (- 2, 2), (- 2, - 2)$

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