Let  $y=f\left(x\right)$ is the solution of the differential equation  $y\frac{d^{2}y}{d{x}^2}-2{\left(\frac{dy}{dx}\right)}^2=0,f\left(1\right)=1.$ If  $S$ is the area bounded by curve  $y=f\left(x\right),y=f^{-1}\left(x\right)$ and  $xy=0,$ in the first quadrant, and if  $f^{-1}\left(2\right)=0,$ then which of the following statement(s) can be TRUE?  $({\log }_{e}2\approx 0.69)$

$$\begin{gathered} y\frac{d^{2}y}{d{x}^2}=2{\left(\frac{dy}{dx}\right)}^2 \\ \frac{\left(\frac{d^{2}y}{d{x}^2}\right)}{\left(\frac{dy}{dx}\right)}=\frac{2\left(\frac{dy}{dx}\right)}{y} \end{gathered}$$

On integrate we get  $\frac{dy}{dx}=k{y}^2$

$$\begin{gathered} \frac{dy}{y^2}=kdx \\ \therefore y=-\frac{1}{kx+c} \end{gathered}$$

Given,  

$$\begin{gathered} f\left(1\right)=1 \\ \therefore c=-1-k\mathrm{.......}\left(i\right) \end{gathered}$$

Given  $f^{-1}\left(2\right)=0\Rightarrow f\left(0\right)=2$

$\therefore 2=-\frac{1}{c}\mathrm{......}\left(ii\right)$

From (i) & (ii) 

$$\begin{gathered} c=k=-\frac{1}{2} \\ \therefore f\left(x\right)=y+\frac{2}{x+1} \\ {f}^{-1}\left(x\right)=\frac{2}{x}-1 \\ S=\underset{0}{\overset{1}{\int }}\frac{2}{x+1}dx+\underset{1}{\overset{2}{\int }}(\frac{2}{x}-1)dx \\ =4\ln 2-1 \end{gathered}$$

Angle of intersection at   $(1,1)$

$$\begin{gathered} f\text{'}\left(x\right)=\frac{-2}{{(x+1)}^2},(f\text{'}\left(x\right))=-\frac{2}{x^2} \\ {m}_1={\left(\frac{dy}{dx}\right)}_{(1,1)}=-\frac{1}{2}{m}_2={\left(\frac{dy}{dx}\right)}_{(1,1)}=2 \\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}angle\hspace{0.17em}}\theta ={\tan }^{-1}\left(\frac{m_1-m_2}{1+m_1{m}_2}\right)={\tan }^{-1}\frac{3}{4} \end{gathered}$$

Similarly we get  $\theta ={\tan }^{-1}\left(\frac{3}{4}\right)at(-2,-2)$