QJEEM08Ma2734: JEE Main 2008 Mathematics - Class 12 JEEM08Ma2734 Solution: The area of the plane region bounded by the curves and is equal …

Question

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The area of the plane region bounded by the curves

x + 2{y^2} = 0

and

\,x + 3{y^2} = 1

is equal to :

Accepted Answer

x + 2{y^2} = 0 \Rightarrow {y^2} = - {x \over 2}

\left[ {} \right.

Left handed parabola with vertex at

\left( {0,0} \right)

\left. {} \right]

x + 3{y^2} = 1 \Rightarrow {y^2} = - {1 \over 3}\left( {x - 1} \right)

\left[ {} \right.

Left handed parabola with vertex at

\left( {1,0} \right)

\left. {} \right]

Solving the two equations we get the points of intersection

as

\left( { - 2,1} \right),\left( { - 2, - 1} \right)

The required area is

ACBDA,

given by

= \left| {\int\limits_{ - 1}^1 {\left( {1 - 3{y^2} - 2{y^2}} \right)dy} } \right|

= \left| {\left[ {y - {{5{y^3}} \over 3}} \right]_{ - 1}^1} \right|

= \left| {\left( {1 - {5 \over 3}} \right) - \left( { - 1 + {5 \over 3}} \right)} \right|

= 2 \times {2 \over 3} = {4 \over 3}\,\,

sq. units.

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