QJEEM19Maa00c: JEE Main 2019 Mathematics - Class 12 JEEM19Maa00c Solution: The area (in sq. units) bounded by the parabolae y = x 2 – 1,…

Question

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The area (in sq. units) bounded by the parabolae y = x² – 1, the tangent at the point (2, 3) to it and the y-axis is :

Accepted Answer

Equation of tangent at (2, 3) on the parabola y = x²

-

1 is

{{y + 3} \over 2} = 2x - 1

\Rightarrow

y + 3 = 4x

-

2

\Rightarrow

y = 4x

-

5

When x = 0 then for the tangent y =

-

5

\therefore

Tangent cuts x y axis at (0,

-

5) point.

\therefore

Area of the bounded region is

=

\int\limits_{ - 5}^3 {{{y + 5} \over 4}} \,\,\,dy - \int\limits_{ - 1}^3 {\sqrt {y + 1} } \,\,\,dy

=

{1 \over 4}\left[ {{{{y^2}} \over 2} + 5y} \right]_{ - 5}^3 - \left[ {{2 \over 3} \times {{\left( {y + 1} \right)}^{{3 \over 2}}}} \right]_{ - 1}^3

{1 \over 4}\left[ {\left( {{9 \over 2} + 15} \right) - \left( {{{25} \over 2} - 25} \right)} \right] - {2 \over 3}{\left( 4 \right)^{{3 \over 2}}}

=

{1 \over 4}\left[ {{{93} \over 2} + {{25} \over 2}} \right] - {2 \over 3} \times 8

=

{1 \over 4} \times {{64} \over 2} - {{16} \over 3}

=

8 - {{16} \over 3}

=

{8 \over 3}

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