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Q.
The slope of normal at any point $(x, y), x>0, y>0$ on the curve $y=y(x)$ is given by $\frac{x^{2}}{x y-x^{2} y^{2}-1}$. If the curve passes through the point $(1,1)$, then $e.y(e)$ is equal to
Slope of normal $=\frac{-d x}{d y}=\frac{x^{2}}{x y-x^{2} y^{2}-1}$
$x^{2} y^{2} d x+d x-x y d x=x^{2} d y$
$x^{2} y^{2} d x+d x=x^{2} d y+x y d x$
$x^{2} y^{2} d x+d x=x(x d y+y d x)$
$x^{2} y^{2} d x+d x=x d(x y)$
$\frac{d x}{x}=\frac{d(x y)}{1+x^{2} y^{2}}$
$\ln kx =\tan ^{-1}( xy ) \ldots$ (i)
passes though $(1,1)$
$\ln k =\frac{\pi}{4} \Rightarrow k = e ^{\frac{\pi}{4}}$
equation (i) be becomes
$\frac{\pi}{4}+\ln x=\tan ^{-1}(x y)$
$x y=\tan \left(\frac{\pi}{4}+\ell \ln x\right)$
$x y=\left(\frac{1+\tan (\ell n x)}{1-\tan (\ell n x)}\right) \ldots$(ii)
put $x=e$ in (ii)
$\therefore \text { ey }( e )=\frac{1+\tan 1}{1-\tan 1}$