AIEEE 2008 Solved Paper

x+2y2=0⇒y2=−‌

x

2

[ Left handed parabola with vertex at (0,0) ]
x+3y2=1⇒y2=−‌

1

3

(x−1)
[ Left handed parabola with vertex at (1,0) ]
Solving the two equations we get the points of intersection
as (−2,1),(−2,−1)
The required area is ACBDA, given by
=|

1

∫

−1

(1−3y2−2y2)dy

‌

‌

|
=|[y−‌

5y3

3

]−11

‌

‌

|
=|(1−‌

5

3

)−(−1+‌

5

3

)

‌

‌

|
=2×‌

2

3

=‌

4

3

sq. units.