Let and respectively be the transverse and conjugate axes of the hyperbola . Then the area of the region above the parabola , below the transverse axis and on the right of the conjugate axis is:

Explanation: Solving 1. Standard Form of the Hyperbola Group the terms and complete the square: Dividing by 16: Center: Transverse Axis ( ): The line Conjugate Axis ( ): The line 2. Defined Region We need the area with the following boundaries: Above the parabola: Below the transverse axis: Right of the conjugate axis: 3. Points of Intersection Find where the parabola meets the transverse axis : Since we are to the right of , we consider the interval from to . 4. Setting up the Integral The area is the integral of the "upper curve" minus the "lower curve": 5. Integration Correct Option: (1)

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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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

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