Question

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

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

Answer 1

Answer: (C)

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