Let T and C respectively be the transverse and conjugate axes of the hyperbola $16 x^{2} - y^{2} + 64 x + 4 y + 44 = 0 .$ Then the area of the region above the parabola $x^{2} = y + 4$ below the
transverse axis T and on the right of the conjugate axis C is

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$4 \sqrt{6} - \frac{44}{3}$

$4 \sqrt{6} - \frac{28}{3}$

$4 \sqrt{6} + \frac{28}{3}$

$4 \sqrt{6} + \frac{44}{3}$

answer is B.

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Detailed Solution

Given hyperbola $16 x^{2} - y^{2} + 64 x + 4 y + 44 = 0$
$\Rightarrow \frac{(x + 2)^{2}}{12} - \frac{(y - 2)^{2}}{4^{2}} = 1$
Centre (-2, 2), $a^{2} = 1^{2}, b^{2} = 4^{2}$
Given parabola $x^{2} = y + 4$
Required Area $= \int_{- \sqrt{6}}^{\sqrt{6}} (2 - (x^{2} - 4) d x - \int_{- \sqrt{6}}^{- 2} (2 - (x^{2} - 4)) d x$
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