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 :

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Answer: A. - {{18} \over {19}} {{dy} \over {dx}} = {{x{y^2} + y} \over x} \Rightarrow {{xdy - ydx} \over {{y^2}}} = xdx \Rightarrow - d\left( {{x \over y}} ight) = d\left( {{{{x^2}} \over 2}} ight) \Rightarrow {{ - x} \over y} = {{{x^2}} \over 2} + C Curve intersect the line x + 2y = 4 at x = - 2 So, - 2 + 2y = 4 \Rightarrow y = 3 So the curve passes through (-2, 3) \Rightarrow {2 \over 3} = 2 + C \Rightarrow C = {{ - 4} \over 3} 	herefore curve is {{ - x} \over y} = {{{x^2}} \over 2} - {4 \over 3} It also passes through (3, y) {{ - 3} \over y} = {9 \over 2} - {4 \over 3} \Rightarrow {{ - 3} \over y} = {{19} \over 6} \Rightarrow y = - {{18} \over {19}}

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 :

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