$$ \begin{aligned} & \text { Slope of normal }=\frac{-d x}{d y}=\frac{x^2}{x y-x^2 y^2-1} \\ & x^2 y^2 d x+d x-x y d x=x^2 d y \\ & x^2 y^2 d x+d x=x^2 d y+x y d x \\ & x^2 y^2 d x+d x=x(x d x+x d x) \\ & x^2 y^2 d x+d x=x d(x y) \\ & \frac{d x}{x}=\frac{d(x y)}{1+x^2 y^2} \\ & \text { In } b x=\tan ^{-1}(x y) \end{aligned} $$ passes though $(1,1)$ $$ \ln k=\frac{\pi}{4} \Rightarrow k=e^{\frac{\pi}{4}} $$ equation (i) be becomes $$ \begin{aligned} & \frac{\pi}{4}+\ln x=\tan ^{-1}(x y) \\ & x y=\tan \left(\frac{\pi}{4}+\ln x\right) \\ & x y=\frac{1+\tan (\ln x)}{1-\tan (\ln x)} \end{aligned} $$ put $\mathrm{x}=\mathrm{e}$ in (ii) $$ \text { ∴ ey }(\mathrm{e})=\frac{1+\tan 1}{1-\tan 1} $$