Question

If the area bounded by the curves $y=a{x}^2$ and $x=a{y}^2$, $(a>0)$ is $1$ sq.units, then the value of $a$ is

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

Answer 1

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

Points of intersection of $y=a{x}^2$ and $x=a{y}^2$ are $(0,0)$ and $(\frac{1}{a},\frac{1}{a})$.

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

$\Rightarrow \frac{2x^{\frac{3}{2}}}{3\sqrt{a}}{|}_0^{\frac{1}{a}}-\frac{a{x}^3}{3}{|}_0^{\frac{1}{a}}=1$

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

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

$\Rightarrow a=\frac{1}{\sqrt{3}}$ ....As $(a>0)$