Q19 of 21 Page 8

Find the area enclosed between the parabola 4y = 3x2 and the straight line 3x – 2y + 12 = 0. [CBSE 2017]

Given curve is

4y = 3x2


and the equation of line is


3x - 2y + 12 = 0



∴ point of intersection is given by solving above two curves-

So, putting y = 3x2/4, we get,

3x - (3/2)x2 + 12 = 0


⇒ 6x - 3x2 + 24 = 0


⇒ 3x2 - 6x - 24 = 0


⇒ 3(x2 -2x - 8) = 0


⇒ (x2 -2x - 8) = 0


⇒ x2 - 4x + 2x - 8 = 0


⇒ x(x-4) + 2(x-4) = 0


⇒ (x+2)(x-4) = 0


∴ x = -2, x = 4

At x = -2, y = 3 and at x = 4, y = 12
So, the points of intersection are (-2, 3) and (4, 12).

Thus, required area





= 45 - 18


= 27 sq. units

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