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MCQMediumJEE Main 2023 · 25 January, Shift 2Conic Sections (Parabola, Ellipse, Hyperbola)

Mathematics Question from JEE Main 2023 · 25 January, Shift 2

Let TT and CC respectively be the transverse and conjugate axes of the hyperbola 16x2−y2+64x+4y+44=016x^2 - y^2 + 64x+ 4y + 44 = 0. Then the area of the region above the parabola x2=y+4x^2 = y + 4, below the transverse axis TT and on the right of the conjugate axis CC is:

  • A

    46+4434\sqrt{6} + \frac{44}{3}

  • B

    46+2834\sqrt{6} + \frac{28}{3}

  • C

    46−4434\sqrt{6} - \frac{44}{3}

  • D

    46−2834\sqrt{6} - \frac{28}{3}

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