The slope of normal at any point $(x, y), x>0, y>0$ on the curve y=y(x) is given by $\frac{x^2}{xy-x^2{y}^2-1}$. If the curve passes through the point (1,1), then $e·y\left(e\right)$ is equal to
 

$$\begin{gathered} slope of normal =\frac{-dx}{dy}=\frac{x^2}{xy-x^2{y}^2-1} \\ \begin{array}{r}{x}^2{y}^{2}dx+dx-xydx=x^{2}dy\\ x^2{y}^{2}dx+dx=x^{2}dy+xydx\end{array} \\ \begin{array}{r}{x}^2{y}^{2}dx+dx=x(xdy+ydx)\\ x^2{y}^{2}dx+dx=xd\left(xy\right)\\ \frac{dx}{x}=\frac{d\left(xy\right)}{1+x^2{y}^2}\\ \ln kx={\tan }^{-1}\left(xy\right)\dots \text{ (i) }\\ \text{ passes though }(1,1)\\ \ln \mathrm{k}=\frac{\pi }{4}\Rightarrow \mathrm{k}={\mathrm{e}}^{\frac{\pi }{4}}\\ \begin{array}{r}\frac{\pi }{4}+\ln x={\tan }^{-1}\left(xy\right)\\ \begin{array}{rl} & xy=\tan \left(\frac{\pi }{4}+\ell n\mathrm{x}\right)\\ & xy=\left(\frac{1+\tan \left(\ell nx\right)}{1-\tan \left(\ell nx\right)}\right)\dots \end{array}\end{array}\\ Put x=e\\ \therefore \text{ e}\left(\text{y }\left(\mathrm{e}\right)\right)=\frac{1+\tan 1}{1-\tan 1}\end{array} \end{gathered}$$