What is the area of the region bounded by the parabolas $y^2=6(x-1)$ and $y^2=3 x ?$

• A

  $\frac{\sqrt{6}}{3}$
• B

  $\frac{2 \sqrt{6}}{3}$
• Correct answer
  C

  $\frac{4 \sqrt{6}}{3}$
• D

  $\frac{5 \sqrt{6}}{3}$

Solution

Given parabolas are $y^2=6(x-1) \quad \ldots .(1)$ and $y^2=3 x$....(2)
$\begin{aligned} & y^2=6(x-1) \\ & \Rightarrow 6 x-6=y^2 \\ & \Rightarrow x=\frac{y^2+6}{6}=\frac{y^2}{6}+1 \end{aligned}$ Also, $\mathrm{y}^2=3 \mathrm{x} \Rightarrow \mathrm{x}=\frac{\mathrm{y}^2}{3}$. Solving (1), (2), $3 \mathrm{x}=6(\mathrm{x}-1)$ $\Rightarrow 3 \mathrm{x}=6 \mathrm{x}-6$ $\Rightarrow 3 \mathrm{x}=6 \Rightarrow \mathrm{x}=2$. $y^2=6 \Rightarrow y= \pm \sqrt{6}$. $\therefore \quad$ Area $\begin{aligned} & =\int_{-\sqrt{6}}^{\sqrt{6}}\left(\frac{y^2}{6}+1-\frac{y^2}{3}\right) \cdot d y=2 \int_0^{\sqrt{6}}\left(1-\frac{y^2}{6}\right) d y=2\left[y-\frac{y^3}{18}\right]_0^{\sqrt{6}} \\ & =2\left[\sqrt{6}-\frac{(\sqrt{6})^3}{18}\right]=2 \times \frac{2 \sqrt{6}}{3}=\frac{4 \sqrt{6}}{3} \end{aligned}$

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