Area of the region enclosed between the curves $y^2=4(x+7)$ and $y^2=5(2-x)$ is

• A

  $\frac{32 \sqrt{2}}{3}$
• B

  $\frac{8}{3}$
• C

  $\frac{1}{6}$
• Correct answer
  D

  $24 \sqrt{5}$

Solution

$\begin{aligned} & 4 x+28=10-5 x \Rightarrow 9 x=-18 \Rightarrow x=-2 \\ & y^2=20 \Rightarrow y=\sqrt{20} \end{aligned}$ Point of intersection is $(-2, \sqrt{20})$ and $(-2,-\sqrt{20})$ Upper curve $\left(2-\frac{y^2}{5}\right) ;$ Lower curve $=\left(\frac{y^2}{4}-7\right)$ Required area is $\int_{-\sqrt{20}}^{\sqrt{20}}\left[\left(2-\frac{y^2}{5}\right)-\left(\frac{y^2}{4}-7\right)\right] d y$ $\begin{aligned} & =\int_{-\sqrt{20}}^{\sqrt{20}}\left(9-\frac{9 y^2}{20}\right) d y=\left[9 y-\frac{3 y^3}{20}\right]_{-\sqrt{20}}^{\sqrt{20}} \\ & =18 \sqrt{20}-\frac{3}{20} \times 40 \sqrt{20}=24 \sqrt{5} \end{aligned}$

Hence, the correct answer is option: 'D'.