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QJEEM20Ma9299: JEE Main 2020 Mathematics - Class 12 JEEM20Ma9299 Solution: Area (in sq. units) of the region outside and inside the…

Area (in sq. units) of the region outside

∣x∣2+∣y∣3=1{{\left| x \right|} \over 2} + {{\left| y \right|} \over 3} = 1 and inside the ellipse x24+y29=1{{{x^2}} \over 4} + {{{y^2}} \over 9} = 1 is :

Multiple Choice Options

  • A 6(4−π)6\left( {4 - \pi } \right)
  • B 3(4−π)3\left( {4 - \pi } \right)
  • C 6(π−2)6\left( {\pi - 2} \right)
  • D 3(π−2)3\left( {\pi - 2} \right)

Step-by-Step Verified Solution

JEE Main 2020 (Online) 2nd September Morning Slot Mathematics - Area Under The Curves Question 103 English Explanation

Area of Ellipse = π\pi ab = 6 π\pi

∴\therefore Required area = Area of ellipse – 4 (Area of triangle OAB)

= 6 π\pi - 4(12×2×3)4\left( {{1 \over 2} \times 2 \times 3} \right)

= 6 π\pi - 12

= 6(π−2)6\left( {\pi - 2} \right) sq.units

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