Question

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

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

Answer 1

Answer: (A)

$\frac{1}{\sqrt{3}}$

$y=a(a{y}^2{)}^{2}y=a^3{y}^{4}y(y^3{a}^3-1)=0$

$y=0,1/a$

$x=a{y}^{2}x=a/a^2$

$x=1/a$

$A=[{\int }_0^{1/a}\sqrt{\frac{x}{a}}-a{x}^{2}dxA={[\frac{2x^{3/2}}{3\sqrt{a}}-\frac{a{x}^3}{3}]}_0^{1/a}A=\frac{2{\left(1/a\right)}^{3/2}}{3\sqrt{a}}-\frac{a{\left(1/a\right)}^3}{3}A=\frac{2}{3a^2}-\frac{1}{3a^2}A=\frac{1}{3a^2}$

given $A=1\frac{1}{3a^2}=13a^2=1a^2=1/3a=\sqrt{1/3}$