JEE Main 2022MathematicsArea Under CurvesMediumMCQ

JEE Main 2022 — Area Under Curves Question with Solution

JEE Main 2022 (27 Jul Shift 1)

Question

The area of the smaller region enclosed by the curves y2=8x+4 and x2+y2+43x-4=0 is equal to

Choose an option

Show full solutionCorrect option: C
Correct answer
C134-123+8π

Step-by-step explanation

Plotting the area of the smaller region enclosed by the curves y2=8x+4 and x2+y2+43x-4=0 we get,

Now finding point of intersection of x2+y2+43x-4=0 and 

y2=8x+4

We get, 0,2 and 0,-2

Both are symmetric about x-axis

So, area will be =2∫0216-y2-23-y2-48dy

=212y16-y2+16sin-1y4-23y-y324+12y02

=138π+4-123

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About this question

This is a previous-year question from JEE Main 2022, covering the Area Under Curves chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.