The slope of the tangent to curve is `(xy^2+y)/x` and it intersects the line `x+2y=4` at `x=-2. If (3,y)` lies on the curve then y is

Options

A. `-18/19`
B. 119
C. `-18/29`
D. None of these

Answer

To solve the problem, we need to find the value of \( y \) such that the point \( (3, y) \) lies on the curve defined by the slope of the tangent, which is given as \( \frac{xy^2 + y}{x} \). We also know that this curve intersects the line \( x + 2y = 4 \) at \( x = -2 \).

### Step-by-Step Solution:

1. **Understanding the Slope of the Tangent:**
The slope of the tangent to the curve is given by:
\[ \frac{dy}{dx} = \frac{xy^2 + y}{x} \]

2. **Rearranging the Slope Equation:**
We can rearrange this equation:
\[ x \frac{dy}{dx} = xy^2 + y \]
This can be rewritten as:
\[ x \frac{dy}{dx} - y = xy^2 \]

3. **Separating Variables:**
We can separate variables:
\[ x \frac{dy}{dx} = xy^2 + y \implies \frac{dy}{y^2 + \frac{y}{x}} = \frac{dx}{x} \]

4. **Integrating Both Sides:**
Integrating both sides, we have:
\[ \int \frac{dy}{y^2 + \frac{y}{x}} = \int \frac{dx}{x} \]
This integration will yield:
\[ -\frac{x}{y} = \frac{x^2}{2} + C \]

5. **Finding the Intersection with the Line:**
The line \( x + 2y = 4 \) can be rearranged to find \( y \):
\[ 2y = 4 - x \implies y = \frac{4 - x}{2} \]
At \( x = -2 \):
\[ y = \frac{4 - (-2)}{2} = \frac{6}{2} = 3 \]
Thus, the curve intersects the line at the point \( (-2, 3) \).

6. **Substituting the Intersection Point:**
Now we substitute \( (-2, 3) \) into the curve equation:
\[ -\frac{-2}{3} = \frac{(-2)^2}{2} + C \]
Simplifying gives:
\[ \frac{2}{3} = 2 + C \implies C = \frac{2}{3} - 2 = \frac{2}{3} - \frac{6}{3} = -\frac{4}{3} \]

7. **Final Curve Equation:**
The equation of the curve becomes:
\[ -\frac{x}{y} = \frac{x^2}{2} - \frac{4}{3} \]

8. **Finding \( y \) when \( x = 3 \):**
Substitute \( x = 3 \) into the curve equation:
\[ -\frac{3}{y} = \frac{3^2}{2} - \frac{4}{3} \]
This simplifies to:
\[ -\frac{3}{y} = \frac{9}{2} - \frac{4}{3} \]
Finding a common denominator (6):
\[ -\frac{3}{y} = \frac{27}{6} - \frac{8}{6} = \frac{19}{6} \]
Cross-multiplying gives:
\[ -3 \cdot 6 = 19y \implies -18 = 19y \implies y = -\frac{18}{19} \]

### Conclusion:
The value of \( y \) such that the point \( (3, y) \) lies on the curve is:
\[ \boxed{-\frac{18}{19}} \]