If `ax^2+2hxy+by^2+2gx+2fy+c=0` then find `dy/dx`

Answer

To find \(\frac{dy}{dx}\) for the equation \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\), we will differentiate both sides of the equation with respect to \(x\).

### Step-by-Step Solution:

1. **Differentiate the equation with respect to \(x\)**:
\[ \frac{d}{dx}(ax^2) + \frac{d}{dx}(2hxy) + \frac{d}{dx}(by^2) + \frac{d}{dx}(2gx) + \frac{d}{dx}(2fy) + \frac{d}{dx}(c) = 0 \]

2. **Differentiate each term**:
- For \(ax^2\):
\[ \frac{d}{dx}(ax^2) = 2ax \]
- For \(2hxy\) (using the product rule):
\[ \frac{d}{dx}(2hxy) = 2h\left(x\frac{dy}{dx} + y\right) \]
- For \(by^2\) (using the chain rule):
\[ \frac{d}{dx}(by^2) = 2by\frac{dy}{dx} \]
- For \(2gx\):
\[ \frac{d}{dx}(2gx) = 2g \]
- For \(2fy\) (using the chain rule):
\[ \frac{d}{dx}(2fy) = 2f\frac{dy}{dx} \]
- For \(c\) (constant):
\[ \frac{d}{dx}(c) = 0 \]

3. **Combine the derivatives**:
\[ 2ax + 2h\left(x\frac{dy}{dx} + y\right) + 2by\frac{dy}{dx} + 2g + 2f\frac{dy}{dx} = 0 \]

4. **Rearranging the equation**:
\[ 2ax + 2hy + 2g + (2hx + 2by + 2f)\frac{dy}{dx} = 0 \]

5. **Isolate \(\frac{dy}{dx}\)**:
\[ (2hx + 2by + 2f)\frac{dy}{dx} = - (2ax + 2hy + 2g) \]

6. **Solve for \(\frac{dy}{dx}\)**:
\[ \frac{dy}{dx} = -\frac{2(ax + hy + g)}{2(hx + by + f)} \]

7. **Simplify the expression**:
\[ \frac{dy}{dx} = -\frac{ax + hy + g}{hx + by + f} \]

### Final Answer:
\[ \frac{dy}{dx} = -\frac{ax + hy + g}{hx + by + f} \]