Let T and C respectively be the transverse and conjugate axes of the hyperbola . Then the area of the region above the parabola , below the transverse axis T and on the right of the conjugate axis C is:
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: D
Solution
First, rewrite the hyperbola by completing the square:
16(x²+4x) - (y²-4y) + 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
This is a horizontal hyperbola centered at (-2, 2). The transverse axis T is the horizontal axis: y = 2. The conjugate axis C is the vertical axis: x = -2.
The parabola x² = y+4 passes through (-2, 0) and has vertex at (0,-4). Rewrite as y = x²-4.
The region is: above the parabola (y ≥ x²-4), below T (y ≤ 2), and right of C (x ≥ -2).
Find where the parabola meets y=2: x²-4=2 → x=√6 (taking right side, x ≥ -2, so x = √6).
The area = ∫ from x=-2 to x=√6 of [2 - (x²-4)] dx = ∫₋₂^√6 (6-x²) dx
= [6x - x³/3] from -2 to √6
= (6√6 - 6√6/3) - (6(-2) - (-8)/3)
= (6√6 - 2√6) - (-12 + 8
16(x²+4x) - (y²-4y) + 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
This is a horizontal hyperbola centered at (-2, 2). The transverse axis T is the horizontal axis: y = 2. The conjugate axis C is the vertical axis: x = -2.
The parabola x² = y+4 passes through (-2, 0) and has vertex at (0,-4). Rewrite as y = x²-4.
The region is: above the parabola (y ≥ x²-4), below T (y ≤ 2), and right of C (x ≥ -2).
Find where the parabola meets y=2: x²-4=2 → x=√6 (taking right side, x ≥ -2, so x = √6).
The area = ∫ from x=-2 to x=√6 of [2 - (x²-4)] dx = ∫₋₂^√6 (6-x²) dx
= [6x - x³/3] from -2 to √6
= (6√6 - 6√6/3) - (6(-2) - (-8)/3)
= (6√6 - 2√6) - (-12 + 8
[JEE Main 2023 · 25 Jan · Shift 2]