If `ax^(2)+2hxy+by^(2)+2gx+2fy+c=0`, then show that `(dy)/(dx).(dx)/(dy) = 1`.

Explanation

To solve the problem, we need to show that \(\frac{dy}{dx} \cdot \frac{dx}{dy} = 1\) given the equation:

\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \]

### Step 1: Differentiate the equation with respect to \(x\)

We start by differentiating the entire equation with respect to \(x\):

\[ \frac{d}{dx}(ax^2) + \frac{d}{dx}(2hxy) + \frac{d}{dx}(by^2) + \frac{d}{dx}(2gx) + \frac{d}{dx}(2fy) + \frac{d}{dx}(c) = 0 \]

Calculating each derivative:

1. \(\frac{d}{dx}(ax^2) = 2ax\)
2. \(\frac{d}{dx}(2hxy) = 2hy + 2hx\frac{dy}{dx}\) (using the product rule)
3. \(\frac{d}{dx}(by^2) = 2by\frac{dy}{dx}\)
4. \(\frac{d}{dx}(2gx) = 2g\)
5. \(\frac{d}{dx}(2fy) = 2f\frac{dy}{dx}\)
6. \(\frac{d}{dx}(c) = 0\)

Putting it all together, we have:

\[ 2ax + 2hy + 2hx\frac{dy}{dx} + 2by\frac{dy}{dx} + 2g + 2f\frac{dy}{dx} = 0 \]

### Step 2: Simplify the equation

Now, we can simplify the equation:

\[ 2ax + 2hy + (2hx + 2by + 2f)\frac{dy}{dx} + 2g = 0 \]

Dividing the entire equation by 2:

\[ ax + hy + (hx + by + f)\frac{dy}{dx} + g = 0 \]

### Step 3: Solve for \(\frac{dy}{dx}\)

Rearranging gives:

\[ (hx + by + f)\frac{dy}{dx} = -ax - hy - g \]

Thus, we can express \(\frac{dy}{dx}\):

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

### Step 4: Differentiate the equation with respect to \(y\)

Now, we differentiate the original equation with respect to \(y\):

\[ \frac{d}{dy}(ax^2) + \frac{d}{dy}(2hxy) + \frac{d}{dy}(by^2) + \frac{d}{dy}(2gx) + \frac{d}{dy}(2fy) + \frac{d}{dy}(c) = 0 \]

Calculating each derivative:

1. \(\frac{d}{dy}(ax^2) = 0\) (since \(x\) is treated as a constant)
2. \(\frac{d}{dy}(2hxy) = 2hx + 2hy\frac{dx}{dy}\)
3. \(\frac{d}{dy}(by^2) = 2by\)
4. \(\frac{d}{dy}(2gx) = 0\)
5. \(\frac{d}{dy}(2fy) = 2f\)
6. \(\frac{d}{dy}(c) = 0\)

Putting it all together, we have:

\[ 2hx + 2hy\frac{dx}{dy} + 2by + 2f = 0 \]

### Step 5: Simplify the equation

Now, we can simplify the equation:

\[ 2hy\frac{dx}{dy} = -2hx - 2by - 2f \]

Dividing the entire equation by 2:

\[ hy\frac{dx}{dy} = -hx - by - f \]

### Step 6: Solve for \(\frac{dx}{dy}\)

Rearranging gives:

\[ \frac{dx}{dy} = \frac{-hx - by - f}{hy} \]

### Step 7: Show that \(\frac{dy}{dx} \cdot \frac{dx}{dy} = 1\)

Now, we can multiply \(\frac{dy}{dx}\) and \(\frac{dx}{dy}\):

\[ \frac{dy}{dx} \cdot \frac{dx}{dy} = \left(\frac{-ax - hy - g}{hx + by + f}\right) \cdot \left(\frac{-hx - by - f}{hy}\right) \]

This simplifies to:

\[ \frac{(ax + hy + g)(hx + by + f)}{(hx + by + f)(hy)} \]

Since the numerator and denominator are equal, we have:

\[ \frac{dy}{dx} \cdot \frac{dx}{dy} = 1 \]

Thus, we have shown that:

\[ \frac{dy}{dx} \cdot \frac{dx}{dy} = 1 \]