Let slope of the tangent

Question:

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

1. $\frac{18}{35}$

2. $-\frac{4}{3}$

3. $-\frac{18}{19}$

4. $-\frac{18}{11}$

Correct Option: , 3

Solution:

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{xy}^{2}+\mathrm{y}}{\mathrm{x}}$

$\frac{x d y-y d x}{y^{2}}=x d x$

$-\mathrm{d}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\mathrm{xdx}$

$-\frac{x}{y}=\frac{x^{2}}{2}+c$

$\because$ curve intersects the line $x+2 y=4$ at

$x=-2 \Rightarrow$ point of intersection is $(-2,3)$

$\therefore$ curve passes through $(-2,3)$

$\Rightarrow \frac{2}{3}=2+c \Rightarrow c=-\frac{4}{3}$

$\Rightarrow \frac{-x}{y}=\frac{x^{2}}{2}-\frac{4}{3}$

Now put $(3, \mathrm{y})$

$\Rightarrow \frac{-3}{y}=\frac{19}{6}$

$\Rightarrow y=\frac{-18}{19}$