Q27 of 45 Page 1

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

Given curve is

4y = 3x2


and the equation of line is


3x - 2y + 12 = 0


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


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


Thus, required area





= 45 - 18


= 27 sq. units


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