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JEE Main 2019 · 12 April · Shift II

JEE Main 2019 Mathematics question on Definite Integrals and Areas

If the area (in sq units) bounded by the parabola y2=4λxy^2 = 4\lambda x and the line y=λxy = \lambda x, λ>0\lambda > 0, is 19\frac{1}{9}, then λ\lambda is equal to
  1. A.
    262\sqrt{6}
  2. B.
    4848
  3. C.
    2424
  4. D.
    434\sqrt{3}
Show answer and solution

Correct option: C

Solution

The parabola y² = 4λx and line y = λx intersect where (λx)² = 4λx, giving λ²x² - 4λx = 0, so x = 0 or x = 4/λ.

Area = ∫₀^(4/λ) [√(4λx) - λx] dx = [2√λ · (2/3)x^(3/2) - λx²/2]₀^(4/λ)

At x = 4/λ: (4√λ/3)(4/λ)^(3/2) - λ/2·(16/λ²) = (4√λ/3)·(8/λ^(3/2)) - 8/λ = 32/(3λ) - 8/λ = 32/(3λ) - 24/(3λ) = 8/(3λ)

Setting 8/(3λ) = 1/9: λ = 72/3 = 24.

[JEE Main 2019 · 12 April · Shift II]

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