Paragraph: The area bounded by the parabola $y^2=k x$ and the line $x=k$, where $k>0$, is $\frac{4}{3}$ square units.
Question: What is the value of $k$ ?

• A

  $\frac{1}{2}$
• Correct answer
  B

  $1$
• C

  $\sqrt{2}$
• D

  $2$

Solution

Given, Area bounded $=\frac{4}{3}$ squar $0=2 \int_0^{\mathrm{k}} \mathrm{ydx}$ $\begin{aligned} & \frac{4}{3}=2 \int_0^{\mathrm{k}} \sqrt{\mathrm{k}} \sqrt{\mathrm{x}} \mathrm{dx} \Rightarrow \frac{4}{3}=2 \sqrt{\mathrm{k}} \frac{2}{3}(\mathrm{k})^{\frac{3}{2}} \Rightarrow 1=\mathrm{k}^2 \\ & \Rightarrow \mathrm{k}= \pm 1 \quad(\text { since } \mathrm{k}>0) \\ & \Rightarrow \mathrm{k}=1\end{aligned}$

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