UPSC NDA Math Model Paper 8

Given Curve: y2=2y−x
⇒y2−2y=−x
Adding one both sides,
⇒y2−2y+1=1−x
⇒(y−1)2=1−x
Above curve is equation is Parabola
⇒(y−1)2=−(x−1)
Curve is symmetry about negative x - axis
Vertex:
y−1=0 and x−1=0
∴y=1 and x=1
⇒(x,y)=(1,1)
Point on intersection:
We know that at y-axis; x = 0
⇒(y−1)2=1−0
⇒y−1=±1
⇒y=±1+1
∴y=2 or 0
(x,y)=(0,2) or (0,0) are Point of intersection. Again, (y−1)2=1−x
∴x=1−(y−1)2
Area under curve =

2

∫

0

xdy
=

2

∫

0

(1−(y−1)2)dy
=

2

∫

0

1.dy−

2

∫

0

(y−1)2.dy
=[y]02−[

(y−1)3

3

]02
=(2−0)−

1

3

[(2−1)3−(0−1)3]
=2−

1

3

[1−(−1)]
=2−

1×2

3

=

6−2

3

=

4

3

Sq. unit