Find the slope of tangent to the curve if `ax^2+2hxy+by^2=0`

Answer

To find the slope of the tangent to the curve given by the equation \( ax^2 + 2hxy + by^2 = 0 \), we will differentiate the equation implicitly with respect to \( x \) and solve for \( \frac{dy}{dx} \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ ax^2 + 2hxy + by^2 = 0 \]

2. **Differentiate both sides with respect to \( x \):**
- The derivative of \( ax^2 \) is \( 2ax \).
- For \( 2hxy \), we will use the product rule:
\[ \frac{d}{dx}(2hxy) = 2h \left( x \frac{dy}{dx} + y \right) \]
- The derivative of \( by^2 \) is \( 2by \frac{dy}{dx} \).

So, differentiating the entire equation gives:
\[ 2ax + 2h \left( x \frac{dy}{dx} + y \right) + 2by \frac{dy}{dx} = 0 \]

3. **Rearrange the equation:**
\[ 2ax + 2hy + 2hx \frac{dy}{dx} + 2by \frac{dy}{dx} = 0 \]

4. **Factor out \( \frac{dy}{dx} \):**
\[ 2hx \frac{dy}{dx} + 2by \frac{dy}{dx} = -2ax - 2hy \]
\[ \frac{dy}{dx} (2hx + 2by) = -2ax - 2hy \]

5. **Solve for \( \frac{dy}{dx} \):**
\[ \frac{dy}{dx} = \frac{-2ax - 2hy}{2hx + 2by} \]
Simplifying gives:
\[ \frac{dy}{dx} = \frac{-(ax + hy)}{hx + by} \]

6. **Conclusion:**
The slope of the tangent to the curve \( ax^2 + 2hxy + by^2 = 0 \) is:
\[ \frac{dy}{dx} = \frac{-(ax + hy)}{hx + by} \]