Q. 195.0( 2 Votes )

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

**[CBSE 2017]**

Answer :

Given curve is

4y = 3x^{2}

and the equation of line is

3x - 2y + 12 = 0

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

So, putting y = 3x^{2}/4, we get,

3x - (3/2)x^{2} + 12 = 0

⇒ 6x - 3x^{2} + 24 = 0

⇒ 3x^{2} - 6x - 24 = 0

⇒ 3(x^{2} -2x - 8) = 0

⇒ (x^{2} -2x - 8) = 0

⇒ x^{2} - 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 = 12So, the points of intersection are (-2, 3) and (4, 12).

Thus, required area

= 45 - 18

= 27 sq. units

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