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:

Step-by-step solution

Given, Equation of hyperbola, 16 x 2 + 4 x - y 2 - 4 y + 44 = 0 ⇒ 16 ( x + 2 ) 2 - 64 - ( y - 2 ) 2 + 4 + 44 = 0 ⇒ 16 x + 2 2 - y - 2 2 = 16 ⇒ x + 2 2 1 - y - 2 2 16 = 1 Hence, equation of conjugate axis will be x = - 2 and equation of transverse axis is given by y = 2 , Now plotting the diagram of parabola x 2 = y + 4 and x = - 2   &   y = 2 we get, Now from above diagram, the area of the bounded region is given by, A = ∫ - 2 6 2 - x 2 - 4 d x ⇒ A = ∫ - 2 6 6 - x 2