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JEE Main 2023
Conic Sections
Hyperbola
Hard

Question

Let T and C respectively be the transverse and conjugate axes of the hyperbola 16x2y2+64x+4y+44=016{x^2} - {y^2} + 64x + 4y + 44 = 0. Then the area of the region above the parabola x2=y+4{x^2} = y + 4, below the transverse axis T and on the right of the conjugate axis C is :

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Solution

This problem requires a strong understanding of conic sections (hyperbola and parabola) and the application of definite integration to calculate the area of a region. We will break down the solution into several key steps: first, analyzing the hyperbola to find its axes; second, analyzing the parabola; third, sketching the region and determining the limits of integration; and finally, evaluating the definite integral.


1. Key Concepts and Formulas

The core concept here is finding the area between two curves using definite integration. If a region is bounded by y=f(x)y=f(x) from above and y=g(x)y=g(x) from below, between x=ax=a and x=bx=b, its area is given by:

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