Question

Find the area enclosed by the parabola $4y=3x^2$ and the line $2y=3x+12$.

• A. $27$ sq. units
• B. $28$ sq. units
• C. $54$ sq. units
• D. $30$ sq. units

Answer 1

Answer: (A)

The given parabola is $y=\frac{3}{4}{x}^2\dots \left(i\right)$
and the line is $3x-2y+12=0\dots \left(ii\right)$
Solving $(1)$ & $(2)$, we get
$3x-2\left(\frac{3}{4}{x}^2\right)+12=0$
$\Rightarrow 3x-\frac{3}{2}{x}^2+12=0$
$\Rightarrow 6x-3x^2+24=0$
$\Rightarrow x^2-2x-8=0$
$\Rightarrow \left(x-4\right)\left(x+2\right)=0$
$\Rightarrow x=4,-2$.
Putting values of $x$ in $\left(1\right)$, we get
$y=\frac{3}{4}{\left(4\right)}^2=12$ and $y=\frac{3}{4}{\left(-2\right)}^2=3$.
Hence, the line and parabola intersects at the points $\left(-2,3\right)$ and $\left(4,12\right)$.

$\therefore$ Required area = area $(ABCDA)$
$=\underset{-2}{\overset{4}{\int }}\left(\frac{3x+12}{2}\right)dx-\left(\underset{-2}{\overset{0}{\int }}\frac{3}{4}{x}^{2}dx+\underset{0}{\overset{4}{\int }}\frac{3}{4}{x}^{2}dx\right)$

$=\frac{1}{2}{\left[\frac{3x^2}{2}+12x\right]}_{-2}^4-\left[\frac{3}{4}{\left[\frac{x^3}{3}\right]}_{-2}^0+\frac{3}{4}{\left[\frac{x^3}{3}\right]}_0^4\right]$

$=\frac{1}{2}\left[\left(24+48\right)-\left(6-24\right)\right]-\left[\frac{3}{4}\left(0+\frac{8}{3}\right)+\frac{3}{4}\left(\frac{64}{3}-0\right)\right]$

$=\frac{1}{2}\left[72+18\right]-\left[2+16\right]=45-18$
$=27$ sq. units