Let T and C respectively be the transverse and conjugate axes of the hyperbola $16{x^2} - {y^2} + 64x + 4y + 44 = 0$. 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 :

Question

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

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

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

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

Solution

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