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Area Under the Curves question

2020 · 2 Sep · Shift 1 · Q21
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Area Under the Curves question

2020 · 2 Sep · Shift 1 · Q21

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
Area (in sq. units) of the region outside ∣x∣2+∣y∣3=1{{\left| x \right|} \over 2} + {{\left| y \right|} \over 3} = 12∣x∣​+3∣y∣​=1 and inside the ellipse x24+y29=1{{{x^2}} \over 4} + {{{y^2}} \over 9} = 14x2​+9y2​=1 is :
  1. A
    6(4−π)6\left( {4 - \pi } \right)6(4−π)
  2. B
    3(4−π)3\left( {4 - \pi } \right)3(4−π)
  3. C
    6(π−2)6\left( {\pi - 2} \right)6(π−2)
  4. D
    3(π−2)3\left( {\pi - 2} \right)3(π−2)
View written solutionFree

Correct answer: C

  1. Identify the two curves

    The line-based curve is ∣x∣2+∣y∣3=1\frac{|x|}{2}+\frac{|y|}{3}=12∣x∣​+3∣y∣​=1 This represents a rhombus with vertices: (±2,0),(0,±3)(\pm 2,0),\quad (0,\pm 3)(±2,0),(0,±3)

    The ellipse is x24+y29=1\frac{x^2}{4}+\frac{y^2}{9}=14x2​+9y2​=1 which has semi-axes 222 and 333.

  2. Required region

    We need the area inside the ellipse and outside the rhombus.

    So, Required area=Area of ellipse−Area of rhombus\text{Required area} = \text{Area of ellipse} - \text{Area of rhombus}Required area=Area of ellipse−Area of rhombus

  3. Area of the ellipse

    For ellipse x2a2+y2b2=1,\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a2x2​+b2y2​=1, area is πab\pi abπab

    Here, a=2a=2a=2 and b=3b=3b=3, so Area of ellipse=π⋅2⋅3=6π\text{Area of ellipse} = \pi \cdot 2 \cdot 3 = 6\piArea of ellipse=π⋅2⋅3=6π

  4. Area of the rhombus

    Its diagonals are:

    • horizontal diagonal =4=4=4
    • vertical diagonal =6=6=6

    Area of rhombus: 12d1d2=12⋅4⋅6=12\frac{1}{2} d_1 d_2 = \frac{1}{2}\cdot 4 \cdot 6 = 1221​d1​d2​=21​⋅4⋅6=12

  5. Required area

    Required area=6π−12=6(π−2)\text{Required area} = 6\pi - 12 = 6(\pi-2)Required area=6π−12=6(π−2)

  6. Match with options

    6(π−2)6(\pi-2)6(π−2) corresponds to Option C.

  7. Verification with stored answer

    Stored correct answer: C

    Our derived answer is also C, so they agree.

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