JEE MAIN - Mathematics (2019 - 9th January Morning Slot - No. 21)
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 :
$$56\over3$$
$$32\over3$$
$$8\over3$$
$$14\over3$$
Explanation
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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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