Question

The ratio in which the area bounded by the curves $y^2=x$ and $x^2=y$ is divided by the line $x=\frac{1}{2}$ is

• A. $\frac{4\sqrt{2}-1}{9-4\sqrt{2}}$
• B. $\frac{3\sqrt{2}+3}{9-4\sqrt{2}}$
• C. $\frac{\sqrt{2}-1}{\sqrt{3}-1}$
• D. $\frac{2\sqrt{2}-1}{3\sqrt{3}-1}$

Answer 1

The correct option is A $\frac{4\sqrt{2}-1}{9-4\sqrt{2}}$

$y^2=x\left(1\right)$

$x^2=y\left(2\right)$

solving (1) & (2)

$y^2-x^2=x-y$

$\Rightarrow \left(y-x\right)\left(y+x\right)-\left(x-y\right)=0$

$\Rightarrow \left(y-x\right)\left(x+y+1\right)=0$

$\Rightarrow y=xorx+y=-1$

$\Rightarrow y=x\left(x+ycannotbenegative\right)$

from (1)

$y^2=x$

$\Rightarrow y^2=y$

$\Rightarrow y^2-y=0$

$\Rightarrow y\left(y-1\right)=0$

$\Rightarrow y=0,1$

so , $x=0,1$

thus points of intersection of given curves are $\left(0,0\right)$ and $\left(1,1\right)$

Ratio of areas $=\frac{{\int }_0^{\frac{1}{2}}\left(y_1-y_2\right)dx}{{\int }_{\frac{1}{2}}^1\left(y_1-y_2\right)dx}$

$=\frac{{\int }_0^{\frac{1}{2}}\left(\sqrt{x}-x^2\right)dx}{{\int }_{\frac{1}{2}}^1\left(\sqrt{x}-x^2\right)dx}$

$=\frac{{{\left[\frac{2}{3}{x}^{\frac{3}{2}}-\frac{x^3}{3}\right]}_0}^{\frac{1}{2}}}{{{\left(\frac{2}{3}{x}^{\frac{3}{2}}-\frac{x^3}{3}\right)}_{\frac{1}{2}}}^1}$

$=\frac{\left(\frac{8-\sqrt{2}}{24\sqrt{2}}\right)}{\frac{1}{3}-\frac{\left(8-\sqrt{2}\right)}{24\sqrt{2}}}$

$=\frac{\left(\frac{8-\sqrt{2}}{24\sqrt{2}}\right)}{\left(\frac{9\sqrt{2}-8}{24\sqrt{2}}\right)}$

$=\frac{8-\sqrt{2}}{9\sqrt{2}-8}$

$=\frac{2\times 4-\sqrt{2}}{9\sqrt{2}-2\times 4}$

$=\frac{4\sqrt{2}-1}{9-4\sqrt{2}}$