Question

Let T and C respectively be the transverse and conjugate axes of the hyperbola 16x² – y² + 64x + 4y + 44 = 0. Then the area of the region above the parabola x² = y + 4, below the transverse axis T and on the right of the conjugate axis C is:

(1) \(4\sqrt 6 + \frac{44}3\)

(2) \(4\sqrt 6 + \frac{28}3\)

(3) \(4\sqrt 6 - \frac{44}3\)

(4) \(4\sqrt 6 - \frac{28}3\)

Answer 1

Correct option is (2) \(4\sqrt 6 + \frac{28}3\)