QJEEM09Mab540: JEE Main 2009 Mathematics - Class 12 JEEM09Mab540 Solution: The area of the region bounded by the parabola the tangent of…

The given parabola is ${\left( {y - 2} \right)^2} = x - 1$

Vertex $\left( {1,2} \right)$ and it meets $x$ -axis at $\left( {5,0} \right)$

Also it gives ${y^2} - 4y - x + 5 = 0$

So, that equation of tangent to the parabola at $\left( {2,3} \right)$ is

$y.3 - 2\left( {y + 3} \right) - {1 \over 2}\left( {x + 2} \right) + 5 = 0$

or $x - 2y + 4 = 0$

which meets $x$ -axis at $\left( { - 4,0} \right).$

In the figure shaded area is the required area.

Let us draw $PD$ perpendicular to $y$ -axis.



Then required area

$= Ar\,\,\Delta BOA + Ar\,\,\left( {OCPD} \right)\, - \,Ar\left( {\Delta APD} \right)$

$= {1 \over 2} \times 4 \times 2 + \int_0^3 {xdy - {1 \over 2}} \times 2 \times 1$

$= 3 + \int_0^3 {{{\left( {y - 2} \right)}^2} + 1\,dy}$

$= 3 + \left[ {{{{{\left( {y - 2} \right)}^3}} \over 3} + y} \right]_0^3$

$= 3 + \left[ {{1 \over 3} + 3 + {8 \over 3}} \right] = 3 + 6 = 9$ Sq. units