# Let slope of the tangent line to a curve at any

Question:

Let slope of the tangent line to a curve at any point $\mathrm{P}(\mathrm{x}, \mathrm{y})$ be given by $\frac{x y^{2}+y}{x}$, 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. (1) $-\frac{18}{11}$

2. (2) $-\frac{18}{19}$

3. (3) $-\frac{4}{3}$

4. (4) $\frac{18}{35}$

Correct Option: , 2

Solution:

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

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

$\Rightarrow-d\left(\frac{x}{y}\right)=d\left(\frac{x^{2}}{2}\right)$

$\Rightarrow \frac{-x}{y}=\frac{z^{2}}{2}+C$

Curve intersect the line $x+2 y=4$ at $x=-2$ So, $-2+2 y=4 \Rightarrow y=3$

So the curve passes through $(-2,3) \Rightarrow \frac{2}{3}=2+C$

$\Rightarrow C=\frac{-4}{3}$

$\therefore$ curve is $\frac{-x}{y}=\frac{x^{2}}{2}-\frac{4}{3}$

It also passes through $(3, y) \frac{-3}{y}=\frac{9}{2}-\frac{4}{3}$

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

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

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