The slope of normal at any point (x, y), x 0, y 0 on the curve y = y(x) is given by [x…

Question

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

Accepted Answer

$$\because$$ $$ - {{dx} \over {dy}} = {{{x^2}} \over {xy - {x^2}{y^2} - 1}}$$

$$\therefore$$ $${{dy} \over {dx}} = {{{x^2}{y^2} - xy + 1} \over {{x^2}}}$$

Let $$xy = v \Rightarrow y + x{{dy} \over {dx}} = {{dv} \over {dx}}$$

$$\therefore$$ $${{dv} \over {dx}} - y = {{({v^2} - v + 1)y} \over v}$$

$$\therefore$$ $${{dv} \over {dx}} = {{{v^2} + 1} \over x}$$

$$\because$$ $$y(1) = 1 \Rightarrow {\tan ^{ - 1}}(xy) = \ln x + {\tan ^{ - 1}}(1)$$

Put $x = e$ and $y = y(e)$ we get

$${\tan ^{ - 1}}(e\,.\,y(e)) = 1 + {\tan ^{ - 1}}1$$

$${\tan ^{ - 1}}(e\,.\,y(e)) - {\tan ^{ - 1}}1 = 1$$

$$\therefore$$ $$e(y(e)) = {{1 + \tan (1)} \over {1 - \tan (1)}}$$

Solutions · 1