If `a x^2+2\ h x y+b y^2+2\ gx+2\ fy+c=0` , find `(dy)/(dx)` and `(dx)/(dy)` . Also, show that `(dy)/(dx)dot(dx)/(dy)=1` .

Explanation

To solve the problem, we need to find \(\frac{dy}{dx}\) and \(\frac{dx}{dy}\) from the given equation:

\[ a x^2 + 2h x y + b y^2 + 2g x + 2f y + c = 0 \]

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

We start by differentiating both sides of the equation with respect to \(x\):

\[ \frac{d}{dx}(a x^2 + 2h x y + b y^2 + 2g x + 2f y + c) = 0 \]

Using the product rule and the chain rule, we differentiate each term:

1. \(a x^2 \rightarrow 2a x\)
2. \(2h x y \rightarrow 2h y + 2h x \frac{dy}{dx}\) (using product rule)
3. \(b y^2 \rightarrow 2b y \frac{dy}{dx}\) (using chain rule)
4. \(2g x \rightarrow 2g\)
5. \(2f y \rightarrow 2f \frac{dy}{dx}\) (using chain rule)
6. \(c\) is a constant, so its derivative is \(0\).

Putting it all together, we have:

\[ 2a x + 2h y + 2h x \frac{dy}{dx} + 2b y \frac{dy}{dx} + 2g + 2f \frac{dy}{dx} = 0 \]

### Step 2: Collect terms involving \(\frac{dy}{dx}\)

Now, we will collect all terms involving \(\frac{dy}{dx}\) on one side:

\[ 2h x \frac{dy}{dx} + 2b y \frac{dy}{dx} + 2f \frac{dy}{dx} = - (2a x + 2h y + 2g) \]

Factoring out \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} (2h x + 2b y + 2f) = - (2a x + 2h y + 2g) \]

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

Now, we can solve for \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{- (2a x + 2h y + 2g)}{2h x + 2b y + 2f} \]

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

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

\[ \frac{d}{dy}(a x^2 + 2h x y + b y^2 + 2g x + 2f y + c) = 0 \]

Using the product rule and chain rule again, we differentiate:

1. \(a x^2 \rightarrow 2x \frac{dx}{dy}\)
2. \(2h x y \rightarrow 2h x + 2h y \frac{dx}{dy}\)
3. \(b y^2 \rightarrow 2b y\)
4. \(2g x\) gives \(2g \frac{dx}{dy}\)
5. \(2f y \rightarrow 2f\)
6. \(c\) is constant, so its derivative is \(0\).

Putting it all together, we have:

\[ 2x \frac{dx}{dy} + 2h x + 2h y \frac{dx}{dy} + 2b y + 2g \frac{dx}{dy} + 2f = 0 \]

### Step 5: Collect terms involving \(\frac{dx}{dy}\)

Collecting terms involving \(\frac{dx}{dy}\):

\[ (2x + 2h y + 2g) \frac{dx}{dy} = - (2h x + 2b y + 2f) \]

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

Now we can solve for \(\frac{dx}{dy}\):

\[ \frac{dx}{dy} = \frac{- (2h x + 2b y + 2f)}{2x + 2h y + 2g} \]

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

Now we need to show that:

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

Substituting the values we found:

\[ \frac{dy}{dx} = \frac{- (2a x + 2h y + 2g)}{2h x + 2b y + 2f} \]

\[ \frac{dx}{dy} = \frac{- (2h x + 2b y + 2f)}{2x + 2h y + 2g} \]

Now, multiply these two:

\[ \frac{dy}{dx} \cdot \frac{dx}{dy} = \left(\frac{- (2a x + 2h y + 2g)}{2h x + 2b y + 2f}\right) \cdot \left(\frac{- (2h x + 2b y + 2f)}{2x + 2h y + 2g}\right) \]

The negatives cancel out:

\[ = \frac{(2a x + 2h y + 2g)(2h x + 2b y + 2f)}{(2h x + 2b y + 2f)(2x + 2h y + 2g)} \]

This simplifies to \(1\), confirming that:

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

### Final Answers:
- \(\frac{dy}{dx} = \frac{- (2a x + 2h y + 2g)}{2h x + 2b y + 2f}\)
- \(\frac{dx}{dy} = \frac{- (2h x + 2b y + 2f)}{2x + 2h y + 2g}\)
- \(\frac{dy}{dx} \cdot \frac{dx}{dy} = 1\)