**The area of the region bonded by the curve x2 = 4y and the straight line x = 4y – 2 is (a) 3/8 sq.units … sq.units (c) 7/8 sq.units…**

###### Answers:

(d) We have parabola x^{2} = 4y and the straight line x = 4y-2

Solving we get

x^{2 }= x + 2

⇒ x^{2}-x-2 = 0

⇒ (x-2)(x+1) = 0

⇒ x = -1,2

For x = -1,y = 1

and for x = 2, y = 1

Thus points of intersection are (-1,1/4)

Graphs of parabola x^{2} = 4y and x = 4y – 2 are as shown in the following figure.