Question

The area cut of from the parabola $4y=3x^2$ by the straight line $2y=3x+12$ is

• A. $25$ sq. units
• B. $27$ sq. units
• C. $36$ sq. units
• D. $16$ sq. units

Answer 1

The correct option is B $27$ sq. units

$4y={3x}^2$ and $2y=3x+12$

$4y={3x}^2$

$\Rightarrow 2y=\frac{3}{2}{x}^2$

$\Rightarrow 3x+12=\frac{3}{2}{x}^2$

$\Rightarrow 6x+24=3x^2$

$\Rightarrow 2x+8=x^2$

$\Rightarrow x^2-2x-8=0$

$x=\frac{2\pm \sqrt{4+32}}{2}=\frac{2\pm 6}{2}$

$\therefore$ $x=4,-2$

and $y=12,3$

$\therefore$ the line $2y=3x+12$ intersects the parabola $4y={3x}^2$ at point $(-2,3)$ & $(4,12)$

$\therefore$ the area enclosed between the parabola and line is shown in the figure as the shaded portion.

Area $={\int }_a^b[f\left(x\right)-g\left(x\right)]dx$

$a=-2$ $b=4$ $f\left(x\right)=\frac{3x+12}{2}g\left(x\right)=\frac{{3x}^2}{4}$

$\therefore$ $A={\int }_{-2}^4[\left[\frac{3x+12}{2}\right]-\left(\frac{{3x}^2}{4}\right)]dx={\int }_{-2}^4\frac{3x+12}{2}dx-{\int }_{-2}^4\frac{{3x}^2}{4}dx$

on integrating we get

$A=\frac{1}{2}{[\frac{{3x}^2}{2}+\frac{12x}{4}]}_2^4-\frac{3}{4}{\left[\frac{x^3}{3}\right]}_{-2}^4$

$=\frac{1}{2}[24+48-6+24]-\frac{1}{4}[64+8]$

$=45-18=27$ sq units (option B)