MathematicsMedium57×since 2003Q3946Let T and C respectively be the transverse and conjugate axes of the hyperbola 16x2−y2+64x+4y+44=016{x^2} - {y^2} + 64x + 4y + 44 = 016x2−y2+64x+4y+44=0. Then the area of the region above the parabola x2=y+4{x^2} = y + 4x2=y+4, below the transverse axis T and on the right of the conjugate axis C is :A46−2834\sqrt 6 - {{28} \over 3}46−328B46−4434\sqrt 6 - {{44} \over 3}46−344C46+2834\sqrt 6 + {{28} \over 3}46+328D46+4434\sqrt 6 + {{44} \over 3}46+344Check answerSkip