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 $x+2y=4\text{at}x=-2$, then the value of $y$, for which the point $(3,y)$ lies on the curve, is :

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

Answer 1

The correct option is B $-\frac{18}{19}$

$\frac{dy}{dx}=\frac{x{y}^2+y}{x}$
$\Rightarrow \frac{xdy-ydx}{y^2}=xdx$
$\Rightarrow -d\left(\frac{x}{y}\right)=d\left(\frac{x^2}{2}\right)$
$\Rightarrow -\frac{x}{y}=\frac{x^2}{2}+C$
Curve intersect the line $x+2y=4\text{at}x=-2$
So, $-2+2y=4\Rightarrow y=3$
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 $(3,y)$
$-\frac{3}{y}=\frac{9}{2}-\frac{4}{3}$
$\Rightarrow -\frac{3}{y}=\frac{19}{6}$
$\Rightarrow y=-\frac{18}{19}$