# Let slope of the tangent line to curve any point P(x,y) be given by $\frac{xy^2+y}{x}$. if the curve intersects theline x+2y=4 at x =-2, then the value of y, for which the point (3,y) lies on  Option: 1 $-\frac{4}{3}$ Option: 2 $-\frac{18}{11}$ Option: 3 $\frac{18}{35}$ Option: 4 $-\frac{18}{19}$

$\\\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{xy}^{2}+\mathrm{y}}{\mathrm{x}} \\ \frac{\mathrm{xdy}-\mathrm{ydx}}{\mathrm{y}^{2}}=\mathrm{xdx} \\ -\mathrm{d}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\mathrm{xdx} \\-\frac{x}{y}=\frac{x^2}{2}+c$

$\\\because\text{ curve intersects the line } x+2 y=4 \text{ at } \mathrm{x}=-2\\ \Rightarrow\text{ point of intersection is } (-2,3)\\ \therefore\text{ curve passes through } (-2,3)$

\begin{aligned} &\Rightarrow \frac{2}{3}=2+c \Rightarrow c=-\frac{4}{3}\\ &\Rightarrow \frac{-x}{y}=\frac{x^{2}}{2}-\frac{4}{3}\\ &\text { Now put }(3, \mathrm{y})\\ &\Rightarrow \frac{-3}{y}=\frac{19}{6}\\ &\Rightarrow y=\frac{-18}{19} \end{aligned}

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