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JEE Main 20199 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
9 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2019 · 9 January · Morning Shift · Mathematics · Area Under Curves

The area (in sq. units) bounded by the parabola y = x ^ 2 - 1, the...

The area (in sq. units) bounded by the parabola y=x2−1,y = {x}^{2} - 1, the tangent at the point (2,3)\left(2, 3\right) to it and the yy-axis isThe area (in sq. units) bounded by the parabola y = x ^ 2 - 1, the tangent at the point (2, 3) to it and the y -axis is

Options

  1. A143\dfrac{14}{3}
  2. B83\dfrac{8}{3}
  3. C323\dfrac{32}{3}
  4. D563\dfrac{56}{3}

Correct answer

B. 83\dfrac{8}{3}

Step-by-step solution

We have, dydx=2x\dfrac{dy}{dx} = 2x

The area (in sq. units) bounded by the parabola y = x ^ 2 - 1, the tangent at the point (2, 3) to it and the y -axis is

∴∴ Slope of tangent at \left(2, 3\right)is</p><p>is</p><p>m = 4</p><p></p><p>∴EquationoftangentisEquation of tangent isy - 3 = 4\left(x - 2\right)

∴∴ Point T≡(0,−5)T ≡ \left(0, - 5\right)

∴∴ Area of δPTM=12×8×2=8\delta PTM = \dfrac{1}{2} \times 8 \times 2 = 8

Area of curve PAMP=∫−13y+1dyPAMP = {∫}_{- 1}^{3}\sqrt{y + 1}dy

=23[(y+1)32]−13=163= \dfrac{2}{3}{\left[\left(y + 1\right)\dfrac{3}{2}\right]}_{- 1}^{3} = \dfrac{16}{3}

∴∴ Required shaded area =8−163=83= 8 - \dfrac{16}{3} = \dfrac{8}{3}

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