Find `dy/dx if ax^2+2hxy+by^2+2gx+2fy+c=0`

Answer

To find the derivative \( \frac{dy}{dx} \) for the equation \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \), we will use implicit differentiation. Here’s a step-by-step solution:

### Step 1: Differentiate the equation with respect to \( x \)
We start with the equation:
\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \]

Now, we differentiate each term with respect to \( x \).

1. The derivative of \( ax^2 \) is \( 2ax \).
2. For \( 2hxy \), we use the product rule:
\[ \frac{d}{dx}(2hxy) = 2h \left( x \frac{dy}{dx} + y \cdot 1 \right) = 2hx \frac{dy}{dx} + 2hy \]
3. The derivative of \( by^2 \) is:
\[ \frac{d}{dx}(by^2) = 2by \frac{dy}{dx} \]
4. The derivative of \( 2gx \) is \( 2g \).
5. The derivative of \( 2fy \) is:
\[ \frac{d}{dx}(2fy) = 2f \frac{dy}{dx} \]
6. The derivative of \( c \) (a constant) is \( 0 \).

Putting it all together, we have:
\[ 2ax + (2hx \frac{dy}{dx} + 2hy) + (2by \frac{dy}{dx}) + 2g + 2f \frac{dy}{dx} + 0 = 0 \]

### Step 2: Rearranging the equation
Now, we can rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ 2ax + 2hy + 2g + (2hx + 2by + 2f) \frac{dy}{dx} = 0 \]

### Step 3: Collect terms involving \( \frac{dy}{dx} \)
We can move the terms without \( \frac{dy}{dx} \) to the right side:
\[ (2hx + 2by + 2f) \frac{dy}{dx} = - (2ax + 2hy + 2g) \]

### Step 4: Solve for \( \frac{dy}{dx} \)
Now, divide both sides by \( (2hx + 2by + 2f) \):
\[ \frac{dy}{dx} = -\frac{2ax + 2hy + 2g}{2hx + 2by + 2f} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is given by:
\[ \frac{dy}{dx} = -\frac{2ax + 2hy + 2g}{2hx + 2by + 2f} \]