Question

The area bounded by the curves $y^2=4a^2(x-1)$and lines $x=1andy=4a$ is

• A. $4a^2$ square units
• B. $\frac{16a}{3}$ square units
• C. $\frac{18a}{3}$ square units
• D. None of these

Answer 1

The correct option is B

$\frac{16a}{3}$ square units

Explanation for the correct option:

Find the area of the bounded region:

Given curves $y^2=4a^2(x-1)$, $x=1andy=4a$

Draw a figure using the given data

If $y=4a$

$$\begin{gathered} {\left(4a\right)}^2=4a^2(x-1) \\ \Rightarrow 16a^2=4a^2(x-1) \\ \Rightarrow x-1=4 \\ \Rightarrow x=5 \end{gathered}$$

Let $y_1=4a$

Let

$$\begin{gathered} {y_2}^2=4a^2(x-1) \\ \Rightarrow y_2=\sqrt{4a^2(x-1)} \\ =2a\sqrt{(x-1)} \end{gathered}$$

Thus required area is${\int }_1^5{y}_1-y_2.dx$

$$\begin{gathered} ={\int }_1^5(4a-2a\sqrt{x-1})dx \\ ={\left[4ax-2a\frac{{(x-1)}^{3/2}}{3/2}\right]}_1^5 \\ =\left[4a\times 5-2a\frac{{(5-1)}^{3/2}}{3/2}-4a\left(1\right)+2a\frac{{(1-1)}^{3/2}}{3/2}\right] \\ =\left[20a-\left(\frac{4a}{3}\times 8\right)-4a+0\right] \\ =16a-\frac{32}{3}a \\ =\frac{16a}{3} \end{gathered}$$

Therefore, the area of the bounded region is $\frac{16a}{3}$ square units.

Hence, option (B) is the correct answer.