If `(a+sqrt(2)bcosx)(a-sqrt(2)bcosy)=a^2-b^2`, find `(dx)/(dy)` at `(pi/4,pi/4)`.

Answer

To solve the given problem, we need to differentiate the equation \((a + \sqrt{2}b \cos x)(a - \sqrt{2}b \cos y) = a^2 - b^2\) and find \(\frac{dx}{dy}\) at the point \((\frac{\pi}{4}, \frac{\pi}{4})\).

### Step-by-Step Solution:

1. **Identify the equation:**
\[ (a + \sqrt{2}b \cos x)(a - \sqrt{2}b \cos y) = a^2 - b^2 \]

2. **Differentiate both sides with respect to \(x\):**
Using the product rule on the left-hand side:
\[ \frac{d}{dx}[(a + \sqrt{2}b \cos x)(a - \sqrt{2}b \cos y)] = 0 \]
Let \(u = a + \sqrt{2}b \cos x\) and \(v = a - \sqrt{2}b \cos y\).

Then,
\[ \frac{du}{dx} = -\sqrt{2}b \sin x \]
and
\[ \frac{dv}{dx} = 0 \quad \text{(since \(y\) is treated as a constant with respect to \(x\))} \]

Applying the product rule:
\[ u \frac{dv}{dx} + v \frac{du}{dx} = 0 \]
This simplifies to:
\[ v(-\sqrt{2}b \sin x) = 0 \]

3. **Differentiate with respect to \(y\):**
Now, differentiate with respect to \(y\):
\[ \frac{d}{dy}[(a + \sqrt{2}b \cos x)(a - \sqrt{2}b \cos y)] = 0 \]
Here, we have:
\[ \frac{du}{dy} = 0 \]
and
\[ \frac{dv}{dy} = \sqrt{2}b \sin y \]

Applying the product rule:
\[ u \frac{dv}{dy} + v \frac{du}{dy} = 0 \]
This simplifies to:
\[ u(\sqrt{2}b \sin y) = 0 \]

4. **Evaluate at \((\frac{\pi}{4}, \frac{\pi}{4})\):**
Substitute \(x = \frac{\pi}{4}\) and \(y = \frac{\pi}{4}\):
\[ \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}, \quad \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \]

Substitute these values into \(u\) and \(v\):
\[ u = a + \sqrt{2}b \cdot \frac{1}{\sqrt{2}} = a + b \]
\[ v = a - \sqrt{2}b \cdot \frac{1}{\sqrt{2}} = a - b \]

5. **Set up the equations:**
From the differentiated equations:
\[ (a - b)(-\sqrt{2}b \cdot \frac{1}{\sqrt{2}}) = 0 \quad \Rightarrow \quad (a - b)(-b) = 0 \]
\[ (a + b)(\sqrt{2}b \cdot \frac{1}{\sqrt{2}}) = 0 \quad \Rightarrow \quad (a + b)(b) = 0 \]

6. **Find \(\frac{dy}{dx}\):**
From the above, we can express \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{b(a - b)}{b(a + b)} = \frac{a - b}{a + b} \]

7. **Find \(\frac{dx}{dy}\):**
To find \(\frac{dx}{dy}\), we take the reciprocal:
\[ \frac{dx}{dy} = \frac{a + b}{a - b} \]

### Final Answer:
\[ \frac{dx}{dy} = \frac{a + b}{a - b} \]