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JEE Main 2009
Exam
JEE Main
Year
2009
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2009 · Mathematics · Area Under Curves

The area of the region bounded by the parabola (y-2)^2=x-1 , the...

The area of the region bounded by the parabola (y−2)2=x−1(y-2)^2=x-1, the tangent to the parabola at the point (2,3)(2,3) and the xx-axis isThe area of the region bounded by the parabola (y-2)^2=x-1 , the tangent to the parabola at the point (2,3) and the xThe area of the region bounded by the parabola (y-2)^2=x-1 , the tangent to the parabola at the point (2,3) and the x

Options

  1. A3
  2. B6
  3. C9
  4. D12

Correct answer

C. 9

Step-by-step solution

Equation of tangent at (2,3)(2,3) to (y−2)2=x−1 is S1=0⇒x−2y+4=0\begin{aligned}& (y-2)^2=x-1 \text{ is }S_1=0 \\ & \Rightarrow x-2 y+4=0 \end{aligned}Required Area = Area of △OCB+\triangle O C B+ Area of OAPD - Area of △PCD\triangle P C D=12(4×2)+∫03(y2−4y+5)dy−12(1×2)=4+[y33−2y2+5y]03−1=4−9−18+15−1=28−19=9 sq. units \begin{aligned}& =\frac{1}{2}(4 \times2)+\int_0^3\left(y^2-4 y+5\right) d y-\frac{1}{2}(1 \times2) \\ & =4+\left[\frac{y^3}{3}-2 y²+5 y\right]_0^3-1=4-9-18+15-1 \\ & =28-19=9 \text{ sq. units } \end{aligned}<imgsrc="/api/proxy−image?url=https<img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_main%252FR4JZ6unjbZ0eLecweDYM2Q5_TFikGBm1Ykf9v2IdkM4.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_main%2FR4JZ6unjbZ0eLecweDYM2Q5_TFikGBm1Ykf9v2IdkM4.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_main%2FR4JZ6unjbZ0eLecweDYM2Q5_TFikGBm1Ykf9v2IdkM4.original.fullsize.png?alt=media"> \\ \text{ (or) } \text{ Area }=\int_0^3\left(2 y-4-y²+4 y-5\right) d y=\int_0^3\left(-y^2+6 y-5\right) d y=-\int_0^3(3-y)^2 d y=\left[\frac{(y-3)^3}{3}\right]_03=0^3=\frac{27}{3}=99\text{ sq.units }$
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