The area in square units bounded by the normal at $(1,2)$ to the parabola $y^{2}=4 x, X$-axis and the curve is given by

• Correct answer
  A

  $\frac{10}{3}$
• B

  $\frac{7}{3}$
• C

  $\frac{4}{3}$
• D

  None of these

Solution

We have, $y^{2}=4 x$ Differentiating w.r.t. $x$, we get $$ \begin{aligned} &2 y \frac{d y}{d x}=4 \Rightarrow \frac{d y}{d x}=\frac{2}{y} \\ &\left(\frac{d y}{d x}\right)_{(1,2)}=\frac{2}{2}=1 \end{aligned} $$

Equation of normal to the curve at $(1,2)$ is $$ \begin{aligned} &\begin{aligned} y-y_{1}=& \frac{1}{\left(-\frac{d y}{d x}\right)}\left(x-x_{1}\right) \Rightarrow(y-2)=-\frac{1}{1}(x-1) \end{aligned} \\ &\Rightarrow \quad y-2=-x+1 \Rightarrow x+y=3 \\ &\text { The line } x+y=3 \text { meets the X-axis at } x=3 \\ &\therefore \text { Required area }=\int_{0}^{1} \sqrt{4 x} d x+\int_{1}^{3}(3-x) d x \\ &=2\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{1}+\left[3 x-\frac{x^{2}}{2}\right]_{1}^{3} \\ &=\frac{4}{3}(1)+\left[9-\frac{9}{2}-3+\frac{1}{2}\right]=\frac{4}{3}+\left(\frac{9}{2}-\frac{5}{2}\right) \\ &=\frac{4}{3}+\frac{4}{2}=\frac{4}{3}+2=\frac{10}{3} \text { sq units } \end{aligned} $$

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