Let slope of the tangent line to a curve at any point P(x, y) be given by ${{x{y^2} + y} \over x}$. If the curve intersects the line x + 2y = 4 at x = $-$2, then the value of y, for which the point (3, y) lies on the curve, is :

${{dy} \over {dx}} = {{x{y^2} + y} \over x}$<br><br>$\Rightarrow {{xdy - ydx} \over {{y^2}}} = xdx$<br>$$ \Rightarrow - d\left( {{x \over y}} \right) = d\left( {{{{x^2}} \over 2}} \right)$$<br><br>$\Rightarrow {{ - x} \over y} = {{{x^2}} \over 2} + C$<br><br>Curve intersect the line x + 2y = 4 at x = $-$ 2<br><br>So, $-$ 2 + 2y = 4 $\Rightarrow$ y = 3<br><br>So the curve passes through ($-$2, 3)<br><br>$\Rightarrow {2 \over 3} = 2 + C$<br><br>$\Rightarrow C = {{ - 4} \over 3}$<br><br>$\therefore$ curve is ${{ - x} \over y} = {{{x^2}} \over 2} - {4 \over 3}$<br><br>It also passes through (3, y)<br><br>${{ - 3} \over y} = {9 \over 2} - {4 \over 3}$<br><br>$\Rightarrow {{ - 3} \over y} = {{19} \over 6}$<br><br>$\Rightarrow y = - {{18} \over {19}}$