The slope of the tangent at a point `P(x, y)` on a curve is `(- (y+3)/(x+2))`. If the curve passes through the origin, find the equation of the curve.

Answer

To find the equation of the curve given that the slope of the tangent at a point \( P(x, y) \) is \( -\frac{(y+3)}{(x+2)} \) and that the curve passes through the origin, we can follow these steps:

### Step 1: Write the differential equation
The slope of the tangent is given by:
\[ \frac{dy}{dx} = -\frac{(y + 3)}{(x + 2)} \]

### Step 2: Rearrange the equation
We can rearrange this equation to separate the variables \( y \) and \( x \):
\[ \frac{dy}{y + 3} = -\frac{dx}{x + 2} \]

### Step 3: Integrate both sides
Now, we integrate both sides:
\[ \int \frac{dy}{y + 3} = \int -\frac{dx}{x + 2} \]

The left side integrates to:
\[ \ln |y + 3| + C_1 \]
And the right side integrates to:
\[ -\ln |x + 2| + C_2 \]

Combining the constants of integration, we can write:
\[ \ln |y + 3| = -\ln |x + 2| + C \]

### Step 4: Exponentiate both sides
Exponentiating both sides gives us:
\[ |y + 3| = e^C \cdot \frac{1}{|x + 2|} \]
Let \( k = e^C \), then we can write:
\[ y + 3 = \frac{k}{x + 2} \]

### Step 5: Solve for \( y \)
Rearranging gives us:
\[ y = \frac{k}{x + 2} - 3 \]

### Step 6: Use the initial condition
Since the curve passes through the origin (0, 0), we can substitute \( x = 0 \) and \( y = 0 \):
\[ 0 = \frac{k}{0 + 2} - 3 \]
This simplifies to:
\[ 0 = \frac{k}{2} - 3 \implies \frac{k}{2} = 3 \implies k = 6 \]

### Step 7: Write the final equation of the curve
Substituting \( k \) back into the equation gives:
\[ y = \frac{6}{x + 2} - 3 \]

Thus, the equation of the curve is:
\[ y = \frac{6}{x + 2} - 3 \]