JEE MAIN - Mathematics (2020 - 2nd September Morning Slot - No. 2)
Area (in sq. units) of the region outside
$${{\left| x \right|} \over 2} + {{\left| y \right|} \over 3} = 1$$ and inside the ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$$ is :
$${{\left| x \right|} \over 2} + {{\left| y \right|} \over 3} = 1$$ and inside the ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$$ is :
$$6\left( {4 - \pi } \right)$$
$$3\left( {4 - \pi } \right)$$
$$6\left( {\pi - 2} \right)$$
$$3\left( {\pi - 2} \right)$$
Explanation
_2nd_September_Morning_Slot_en_2_1.png)
Area of Ellipse = $$\pi $$ab = 6$$\pi $$
$$ \therefore $$ Required area = Area of ellipse ā 4 (Area of triangle OAB)
= 6$$\pi $$ - $$4\left( {{1 \over 2} \times 2 \times 3} \right)$$
= 6$$\pi $$ - 12
= $$6\left( {\pi - 2} \right)$$ sq.units
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