If `(a+sqrt2b cosx)(a-sqrt2b cosy)=a^2-b^2, ` where `a gt b gt 0` , then `(dx)/(dy) ` at `(pi/4,pi/4)` is :

Options

A. `(a-b)/(a+b)`
B. `(a+b)/(a-b)`
C. `(2a+b)/(2a-b)`
D. `(a-2b)/(a+2b)`

Answer

To solve 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 \((\frac{\pi}{4}, \frac{\pi}{4})\), we will follow these steps:

### Step 1: Differentiate the equation with respect to \(y\)
We will apply the product rule to differentiate the left-hand side:

\[ \frac{d}{dy}[(a + \sqrt{2} b \cos x)(a - \sqrt{2} b \cos y)] = 0 \]

Using the product rule:

\[ \frac{d}{dy}(u \cdot v) = u \frac{dv}{dy} + v \frac{du}{dy} \]

Let \(u = a + \sqrt{2} b \cos x\) and \(v = a - \sqrt{2} b \cos y\).

Differentiating \(u\) with respect to \(y\):

\[ \frac{du}{dy} = 0 \quad \text{(since \(u\) does not depend on \(y\))} \]

Differentiating \(v\) with respect to \(y\):

\[ \frac{dv}{dy} = \sqrt{2} b \sin y \]

Thus, we have:

\[ 0 = u \cdot \sqrt{2} b \sin y + v \cdot 0 \]

This simplifies to:

\[ 0 = (a + \sqrt{2} b \cos x)(\sqrt{2} b \sin y) \]

### Step 2: Differentiate the right-hand side
The right-hand side \(a^2 - b^2\) is constant with respect to \(y\), so its derivative is:

\[ \frac{d}{dy}(a^2 - b^2) = 0 \]

### Step 3: Set up the equation
From the differentiation, we have:

\[ (a + \sqrt{2} b \cos x)(\sqrt{2} b \sin y) + (a - \sqrt{2} b \cos y)(\sqrt{2} b \sin x) \frac{dx}{dy} = 0 \]

Rearranging gives:

\[ (a + \sqrt{2} b \cos x)(\sqrt{2} b \sin y) = -(a - \sqrt{2} b \cos y)(\sqrt{2} b \sin x) \frac{dx}{dy} \]

### Step 4: Solve for \(\frac{dx}{dy}\)
Now, we can isolate \(\frac{dx}{dy}\):

\[ \frac{dx}{dy} = -\frac{(a + \sqrt{2} b \cos x)(\sqrt{2} b \sin y)}{(a - \sqrt{2} b \cos y)(\sqrt{2} b \sin x)} \]

### Step 5: Evaluate at \((\frac{\pi}{4}, \frac{\pi}{4})\)
Now we substitute \(x = \frac{\pi}{4}\) and \(y = \frac{\pi}{4}\):

\[ \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}, \quad \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \]

Substituting these values into our expression for \(\frac{dx}{dy}\):

\[ \frac{dx}{dy} = -\frac{(a + \sqrt{2} b \cdot \frac{1}{\sqrt{2}})(\sqrt{2} b \cdot \frac{1}{\sqrt{2}})}{(a - \sqrt{2} b \cdot \frac{1}{\sqrt{2}})(\sqrt{2} b \cdot \frac{1}{\sqrt{2}})} \]

This simplifies to:

\[ \frac{dx}{dy} = -\frac{(a + b)(b)}{(a - b)(b)} \]

Canceling \(b\) from numerator and denominator (since \(b > 0\)) gives:

\[ \frac{dx}{dy} = -\frac{a + b}{a - b} \]

### Final Answer
Thus, the value of \(\frac{dx}{dy}\) at \((\frac{\pi}{4}, \frac{\pi}{4})\) is:

\[ \frac{dx}{dy} = -\frac{a + b}{a - b} \]