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Area Under the Curves question

2022 · 27 Jul · Shift 1 · Q32
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  5. /2022 · 27 Jul · Shift 1 · Q32

Area Under the Curves question

2022 · 27 Jul · Shift 1 · Q32

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area of the smaller region enclosed by the curves y2=8x+4y^{2}=8 x+4y2=8x+4 and x2+y2+43x−4=0x^{2}+y^{2}+4 \sqrt{3} x-4=0x2+y2+43​x−4=0 is equal to
  1. A
    13(2−123+8π)\frac{1}{3}(2-12 \sqrt{3}+8 \pi)31​(2−123​+8π)
  2. B
    13(2−123+6π)\frac{1}{3}(2-12 \sqrt{3}+6 \pi)31​(2−123​+6π)
  3. C
    13(4−123+8π)\frac{1}{3}(4-12 \sqrt{3}+8 \pi)31​(4−123​+8π)
  4. D
    13(4−123+6π)\frac{1}{3}(4-12 \sqrt{3}+6 \pi)31​(4−123​+6π)
View written solutionFree

Correct answer: C

  1. Write the curves in standard form

Given: y2=8x+4 ⇒ x=y2−48=y28−12y^2=8x+4 \,\Rightarrow\, x=\frac{y^2-4}{8}=\frac{y^2}{8}-\frac12y2=8x+4⇒x=8y2−4​=8y2​−21​ This is a right-opening parabola.

For the circle: x2+y2+43 x−4=0x^2+y^2+4\sqrt3\,x-4=0x2+y2+43​x−4=0 Complete the square in xxx: x2+43x=(x+23)2−12x^2+4\sqrt3 x=(x+2\sqrt3)^2-12x2+43​x=(x+23​)2−12 So, (x+23)2+y2=16(x+2\sqrt3)^2+y^2=16(x+23​)2+y2=16 Hence it is a circle with center (−23,0)(-2\sqrt3,0)(−23​,0) and radius 444.

  1. Find points of intersection

Substitute x=y2−48x=\dfrac{y^2-4}{8}x=8y2−4​ into the circle: (y2−48)2+y2+43(y2−48)−4=0\left(\frac{y^2-4}{8}\right)^2+y^2+4\sqrt3\left(\frac{y^2-4}{8}\right)-4=0(8y2−4​)2+y2+43​(8y2−4​)−4=0 Multiply by 161616: (y2−4)24+16y2+83(y2−4)−64=0\frac{(y^2-4)^2}{4}+16y^2+8\sqrt3(y^2-4)-64=04(y2−4)2​+16y2+83​(y2−4)−64=0 Multiply by 444: (y2−4)2+64y2+323(y2−4)−256=0(y^2-4)^2+64y^2+32\sqrt3(y^2-4)-256=0(y2−4)2+64y2+323​(y2−4)−256=0 Expand: y4−8y2+16+64y2+323y2−1283−256=0y^4-8y^2+16+64y^2+32\sqrt3 y^2-128\sqrt3-256=0y4−8y2+16+64y2+323​y2−1283​−256=0 y4+(56+323)y2−(240+1283)=0y^4+(56+32\sqrt3)y^2-(240+128\sqrt3)=0y4+(56+323​)y2−(240+1283​)=0 Let t=y2t=y^2t=y2. Then t2+(56+323)t−(240+1283)=0t^2+(56+32\sqrt3)t-(240+128\sqrt3)=0t2+(56+323​)t−(240+1283​)=0 Checking t=4t=4t=4: 16+4(56+323)−(240+1283)=16+224+1283−240−1283=016+4(56+32\sqrt3)-(240+128\sqrt3)=16+224+128\sqrt3-240-128\sqrt3=016+4(56+323​)−(240+1283​)=16+224+1283​−240−1283​=0 So t=4t=4t=4 is a root. Thus y2=4⇒y=±2y^2=4 \Rightarrow y=\pm 2y2=4⇒y=±2 Then from parabola, x=4−48=0x=\frac{4-4}{8}=0x=84−4​=0 So the curves intersect at (0,2), (0,−2).(0,2),\,(0,-2).(0,2),(0,−2).

  1. Decide the smaller enclosed region

Between y=−2y=-2y=−2 and y=2y=2y=2:

  • parabola gives xp=y28−12x_p=\frac{y^2}{8}-\frac12xp​=8y2​−21​
  • circle gives right branch xc=−23+16−y2x_c=-2\sqrt3+\sqrt{16-y^2}xc​=−23​+16−y2​

At y=0y=0y=0: xp=−12,xc=4−23x_p=-\frac12,\qquad x_c=4-2\sqrt3xp​=−21​,xc​=4−23​ Since 4−23≈0.5364-2\sqrt3\approx 0.5364−23​≈0.536, we have xc>xpx_c>x_pxc​>xp​. Thus the smaller enclosed region is the region between the parabola (left boundary) and the right branch of the circle (right boundary), for −2≤y≤2-2\le y\le 2−2≤y≤2.

Hence area is A=∫−22[xc−xp]dyA=\int_{-2}^{2}\left[x_c-x_p\right]dyA=∫−22​[xc​−xp​]dy A=∫−22(−23+16−y2−y28+12)dyA=\int_{-2}^{2}\left(-2\sqrt3+\sqrt{16-y^2}-\frac{y^2}{8}+\frac12\right)dyA=∫−22​(−23​+16−y2​−8y2​+21​)dy

By symmetry, A=2∫02(−23+16−y2−y28+12)dyA=2\int_0^2\left(-2\sqrt3+\sqrt{16-y^2}-\frac{y^2}{8}+\frac12\right)dyA=2∫02​(−23​+16−y2​−8y2​+21​)dy

  1. Evaluate the integrals

So, A=2[∫0216−y2 dy+∫02(12−23)dy−∫02y28 dy]A=2\left[\int_0^2\sqrt{16-y^2}\,dy+\int_0^2\left(\frac12-2\sqrt3\right)dy-\int_0^2\frac{y^2}{8}\,dy\right]A=2[∫02​16−y2​dy+∫02​(21​−23​)dy−∫02​8y2​dy]

(i) Compute ∫0216−y2 dy\int_0^2 \sqrt{16-y^2}\,dy∫02​16−y2​dy

Use the standard formula: ∫a2−y2 dy=y2a2−y2+a22sin⁡−1(ya)\int \sqrt{a^2-y^2}\,dy=\frac y2\sqrt{a^2-y^2}+\frac{a^2}{2}\sin^{-1}\left(\frac ya\right)∫a2−y2​dy=2y​a2−y2​+2a2​sin−1(ay​) Here a=4a=4a=4. Thus

=\left[\frac y2\sqrt{16-y^2}+8\sin^{-1}\left(\frac y4\right)\right]_0^2$$ $$=\frac{2}{2}\sqrt{12}+8\sin^{-1}\left(\frac12\right) =2\sqrt3+8\cdot \frac\pi6 =2\sqrt3+\frac{4\pi}{3}$$ ### (ii) Compute constant term $$\int_0^2\left(\frac12-2\sqrt3\right)dy=2\left(\frac12-2\sqrt3\right)=1-4\sqrt3$$ ### (iii) Compute parabola term $$\int_0^2\frac{y^2}{8}dy=\frac18\cdot \left[\frac{y^3}{3}\right]_0^2=\frac18\cdot \frac83=\frac13$$ Therefore, $$A=2\left(2\sqrt3+\frac{4\pi}{3}+1-4\sqrt3-\frac13\right)$$ $$=2\left(\frac23-2\sqrt3+\frac{4\pi}{3}\right)$$ $$=\frac43-4\sqrt3+\frac{8\pi}{3}$$ Factor out $\frac13$: $$A=\frac13(4-12\sqrt3+8\pi)$$ --- 5. **Match with the options** This equals **Option C**: $$\boxed{\frac13(4-12\sqrt3+8\pi)}$$ --- 6. **Compare with stored correct answer** Stored correct answer: **C** Our derived answer is also **C**, so they agree.
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