Question

The maximum area bounded by the curves $y^2=4ax,y=ax$ and $y=\frac{x}{a}(1<a\le 2)$

• A. $84$
• B. $64$
• C. $50$
• D. $48$

Answer 1

The correct option is A $84$

The curves are $y^2=4ax$ and $y=ax.$ At their point of intersection
$a^2{x}^2=4ax\Rightarrow ax=4,x=0x=\frac{4}{a}\Rightarrow A(\frac{4}{a},4)$

Similarly for $y^2=4ax$ and $y=\frac{x}{a},$
$\frac{x^2}{a^2}=4ax$
$\Rightarrow x=4a^3$
$\Rightarrow B(4a^3,4a^2)$

$\text{Area}\left(OAB\right)=\underset{0}{\overset{4/a}{\int }}(ax-\frac{x}{a})dx+\underset{4/a}{\overset{4a^3}{\int }}(\sqrt{4ax}-\frac{x}{a})dx$

$A=\frac{8}{3}(a^5-\frac{1}{a})$

$\frac{dA}{da}=\frac{8}{3}\cdot (5a^4+\frac{1}{a^2})>0\forall a$
$\Rightarrow A\left(a\right)$ is increasing function
$\therefore A_{max}={\frac{8}{3}(a^5-\frac{1}{a})|}_{a=2}=\frac{8}{3}(32-\frac{1}{2})=84\text{sq. units}$