Question

If the area bounded by the curve $y=a{x}^2$ and $x=a{y}^2,(a>0)$ is $3sq.units$, then the value of $a$ is

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $1$
• D. $4$

Answer 1

The correct option is B $\frac{1}{3}$

The parabolas intersect at $(0,0)$ and $(\frac{1}{a},\frac{1}{a})$

Area bounded $={\int }_0^{\frac{1}{a}}(\sqrt{\frac{x}{a}}-a{x}^2)dx$

$={[\frac{2}{3}\frac{x^{\frac{3}{2}}}{\sqrt{a}}-a\frac{x^3}{3}]}_0^{\frac{1}{a}}$

$=[\frac{2}{3}\frac{1}{\sqrt{a}\left(a^{\frac{3}{2}}\right)}-\frac{a}{3a^3}]$

$=[\frac{2}{3a^2}-\frac{1}{3a^2}]$

$=\frac{1}{3a^2}$

It is given that area $=3$ sq.units

Therefore, $\frac{1}{3a^2}=3$

$⟹a=\frac{1}{3}$

Hence, answer is option (B).