Question

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

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

Answer 1

Answer: (B)

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

Given, $y^2=4a^2(x-1)$ and $y=4a$
Therefore, $16a^2=4a^2(x-1)$
$\Rightarrow x=5$
Thus required area is ${\int }_1^5(4a-2a\sqrt{x-1})dx$
$={[4ax-2a\frac{{(x-1)}^{3/2}}{3/2}]}_1^5$
$=[20a-\frac{4a}{3}\times 8-4a+0]$
$=16a-\frac{32}{3}a=\frac{16}{3}a$ sq units