Question

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

• A. $\frac{18}{35}$
• B. $-\frac{4}{3}$
• C. $-\frac{18}{19}$
• D. $-\frac{18}{11}$

Answer 1

Answer: (C)

$\frac{dy}{dx}=\frac{x{y}^2+y}{x}$
$\frac{xdy-ydx}{y^2}=xdx$
$-d\left(\frac{x}{y}\right)=xdx$
$-\frac{x}{y}=\frac{x^2}{2}+c$
$\because$ curve intersects the line $x+2y=4$ 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 $(3,y)$
$\Rightarrow \frac{-3}{y}=\frac{19}{6}$
$\Rightarrow y=\frac{-18}{19}$