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

<p>$\because$ $- {{dx} \over {dy}} = {{{x^2}} \over {xy - {x^2}{y^2} - 1}}$</p> <p>$\therefore$ ${{dy} \over {dx}} = {{{x^2}{y^2} - xy + 1} \over {{x^2}}}$</p> <p>Let $xy = v \Rightarrow y + x{{dy} \over {dx}} = {{dv} \over {dx}}$</p> <p>$\therefore$ ${{dv} \over {dx}} - y = {{({v^2} - v + 1)y} \over v}$</p> <p>$\therefore$ ${{dv} \over {dx}} = {{{v^2} + 1} \over x}$</p> <p>$\because$ $y(1) = 1 \Rightarrow {\tan ^{ - 1}}(xy) = \ln x + {\tan ^{ - 1}}(1)$</p> <p>Put $x = e$ and $y = y(e)$ we get</p> <p>${\tan ^{ - 1}}(e\,.\,y(e)) = 1 + {\tan ^{ - 1}}1$</p> <p>${\tan ^{ - 1}}(e\,.\,y(e)) - {\tan ^{ - 1}}1 = 1$</p> <p>$\therefore$ $e(y(e)) = {{1 + \tan (1)} \over {1 - \tan (1)}}$</p>