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AIEEE 2009 Solved Paper

Section: Mathematics

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Question : 77 of 89

Marks: +1, -0

The area of the region bounded by the parabola

(y-2)^2=x-1

, the tangent of the parabola at the point

(2,3)

x

6
9
12
3

Solution:

The given parabola is

(y-2)^2=x-1

Vertex

(1,2)

and it meets

x

-axis at

(5,0)

Also it gives

y^2-4y-x+5=0

So, that equation of tangent to the parabola at

(2,3)

is

y .3-2(y+3)-\;\frac{1}{2}(x+2)+5=0

or

x-2y+4=0

which meets

x

-axis at

(-4,0)

.

In the figure shaded area is the required area.

Let us draw

P D

perpendicular to

y

-axis.

Then required area

\;=A r \Delta B O A+A r(O C P D)-A r(\triangle A P D)

\;=\;\frac{1}{2} \times 4 \times 2+\int\limits_{0}^{3} x dy-\;\frac{1}{2} \times 2 \times 1

\;=3+\int\limits_{0}^{3}(y-2)^2+1 dy

\;=3+ \left[ \;\frac{(y-2)^3}{3}+y \right]_{0}^{3}

\;=3+ \left[ \;\frac{1}{3}+3+\;\frac{8}{3} \right] =3+6=9 \;\text{ Sq. units }\;

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