Question

The area enclosed between the curves $y=a{x}^2$ and
$x=a{y}^2(a>0)$ is 1 sq unit. Then, the value of a is

• A. $(a)\frac{1}{\sqrt{3}}$
• B. $(b)\frac{1}{2}$
• C. $(c)1$
• D. $(d)\frac{1}{3}$

Answer 1

Answer: (A)

As from the figure, area enclosed between the curves is
OABCO.
Thus, the point of intersection of
$y=a{x}^2$and $x=a{y}^2$
$\Rightarrow x=a(a{x}^2{)}^2$
$\Rightarrow x=0,\frac{1}{a}\Rightarrow y=0,\frac{1}{a}$
So, the points of intersection are (0, 0) and $(\frac{1}{a},\frac{1}{a})$
$\therefore$ Required area OABCO = Area of curve OCBDO
- Area of curve OABDO
$\Rightarrow \underset{0}{\overset{1/a}{\int }}(\sqrt{\frac{x}{a}}-a{x}^2)dx=1[given]$
$\Rightarrow [\frac{1}{\sqrt{a}}.\frac{x^{3/2}}{3/2}-\frac{a{x}^3}{3}{]}_0^{1/a}=1$
$\Rightarrow \frac{2}{3a^2}-\frac{1}{3a^2}=1$
$\Rightarrow a^2=\frac{1}{3}\Rightarrow a=\frac{1}{\sqrt{3}}[\because a>0]$