JEE Main 2023 · 25 January · evening · Single correct · +4 / −1
Let T and C respectively be the transverse and conjugate axes of the hyperbola 16x² - y² + 64x + 4y + 44 = 0. Then the area of the region above the parabola x² = y + 4, below the transverse axis T and on the right of the conjugate axis C is :
- (a) 4√6 - 28/3
- (b) 4√6 - 44/3
- (c) 4√6 + 28/3
- (d) 4√6 + 44/3
Answer and solution
Answer (c) 4√6 + 28/3
Solution
Maths sir · 5 steps
Complete the squares first, because the hyperbola's axes are just the lines y = 2 and x = −2.
- Standard form. Group the terms: 16(x+2)² - 64 - (y-2)² + 4 + 44 = 0, so 16(x+2)² - (y-2)² = 16. That is (x+2)²/1 - (y-2)²/16 = 1, with centre (-2, 2).
- Find the axes. The x-term is positive, so the transverse axis runs horizontally through the centre: T is y = 2. The conjugate axis is the vertical line through the centre: C is x = -2.
- Set the limits. The parabola is y = x² - 4. It meets y = 2 where x² = 6, so x = ±√6. The region lies to the right of x = -2, so x runs from -2 to √6. In that range the height of the strip is 2 - (x² - 4) = 6 - x².
A = ∫₋₂^(√6) (6 - x²) dx = [6x - x³/3]₋₂^(√6)
- Evaluate. At √6: 6√6 - 6√6/3 = 4√6. At -2: -12 - (-8)/3 = -12 + 8/3 = -28/3.
- So. A = 4√6 - (-28/3) = 4√6 + 28/3, which is option (c).
The idea to keep. For a shifted conic, complete the squares to find the centre. The axes are then the horizontal and vertical lines through it, and these lines become your integration limits.
Answer from the NTA final answer key. Solution by PrepWiser.