Question

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

$-\frac{18}{19}$

Explanation for the correct option:

Step 1:Finding the equation of the curve using information on slope:

Given that the slope of the tangent line is given as,

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

$$\begin{gathered} \Rightarrow xdy=\left(x{y}^2+y\right)dx \\ \Rightarrow xdy=x{y}^{2}dx+ydx \\ \Rightarrow x{y}^{2}dx=xdy-ydx \end{gathered}$$

Now divide $y^2$ on both sides of the above equation.

$$\begin{gathered} \Rightarrow \frac{x{y}^2}{y^2}dx=\frac{xdy-ydx}{y^2} \\ \Rightarrow xdx=\frac{xdy-ydx}{y^2} \end{gathered}$$

Since, the differential of $-d\left(\frac{x}{y}\right)=\frac{xdy-ydx}{y^2}$ then the above equation can be written as,

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

Integrating on both sides we get,

$$\begin{gathered} \Rightarrow \frac{x^2}{2}=-\frac{x}{y}+c \\ \Rightarrow -\frac{x}{y}=\frac{x^2}{2}+c \end{gathered}$$

Since the curve intersects the line $x+2y=4$ at $x=-2$ we get,

$\begin{array}{rcl}& \Rightarrow & -2+2y=4\\ 2y& =& 6\\ y& =& 3\end{array}$

Therefore, the curve intersects the line $x+2y=4$ at $\left(-2,3\right)$.

Step 2: Finding the value of c:

Now substitute $x$ as $-2$ and $y$ as $3$ in the equation of$-\frac{x}{y}=\frac{x^2}{2}+c$.

$\begin{array}{rcl}& \Rightarrow & -\frac{-2}{3}=\frac{{\left(-2\right)}^2}{2}+c\\ \frac{2}{3}& =& \frac{4}{2}+c\\ c& =& -\frac{4}{3}\end{array}$

Substituting the value of $c$ in the equation of$-\frac{x}{y}=\frac{x^2}{2}+c$.

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

Step 3: Finding the value of y:

Now since the point $\left(3,y\right)$ lies on the curve we get,

$\begin{array}{rcl}& \Rightarrow & -\frac{3}{y}=\frac{3^2}{2}-\frac{4}{3}\\ -\frac{3}{y}& =& \frac{9}{2}-\frac{4}{3}\\ -\frac{3}{y}& =& \frac{19}{6}\\ y& =& -\frac{18}{19}\end{array}$

Therefore, the value of $y$ is equal to$-\frac{18}{19}$.

Hence, the correct answer is option (B)