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 x2 = 4y and the straight line x = 4y-2
Solving we get
x2 = x + 2
⇒ x2-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 x2 = 4y and x = 4y – 2 are as shown in the following figure.