If `sqrt(1-x^(2))+sqrt(1-y^(2))=a(x-y)`, then `(dy)/(dx)` equals

Options

A. `sqrt((1-x^(2))(1-y^(2)))`
B. `sqrt((1-y^(2))/(1-x^(2)))`
C. `sqrt((1-x^(2))/(1-y^(2)))`
D. none of these

Answer

To solve the equation \( \sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y) \) and find \( \frac{dy}{dx} \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y) \]

### Step 2: Substitute trigonometric identities
Let \( x = \sin(\theta) \) and \( y = \sin(\phi) \). Then, we can rewrite the equation as:
\[ \sqrt{1 - \sin^2(\theta)} + \sqrt{1 - \sin^2(\phi)} = a(\sin(\theta) - \sin(\phi)) \]
Using the identity \( \sqrt{1 - \sin^2(t)} = \cos(t) \), we have:
\[ \cos(\theta) + \cos(\phi) = a(\sin(\theta) - \sin(\phi)) \]

### Step 3: Simplify the equation
Now, we can express \( a \):
\[ a = \frac{\cos(\theta) + \cos(\phi)}{\sin(\theta) - \sin(\phi)} \]

### Step 4: Use trigonometric identities
Using the sum-to-product identities:
\[ \cos(\theta) + \cos(\phi) = 2 \cos\left(\frac{\theta + \phi}{2}\right) \cos\left(\frac{\theta - \phi}{2}\right) \]
\[ \sin(\theta) - \sin(\phi) = 2 \sin\left(\frac{\theta + \phi}{2}\right) \cos\left(\frac{\theta - \phi}{2}\right) \]
Substituting these into the expression for \( a \):
\[ a = \frac{2 \cos\left(\frac{\theta + \phi}{2}\right) \cos\left(\frac{\theta - \phi}{2}\right)}{2 \sin\left(\frac{\theta + \phi}{2}\right) \cos\left(\frac{\theta - \phi}{2}\right)} \]
The \( 2 \cos\left(\frac{\theta - \phi}{2}\right) \) cancels out:
\[ a = \frac{\cos\left(\frac{\theta + \phi}{2}\right)}{\sin\left(\frac{\theta + \phi}{2}\right)} = \cot\left(\frac{\theta + \phi}{2}\right) \]

### Step 5: Differentiate both sides
Now differentiate both sides of the original equation with respect to \( x \):
\[ \frac{d}{dx}(\sqrt{1 - x^2}) + \frac{d}{dx}(\sqrt{1 - y^2}) = \frac{d}{dx}(a(x - y)) \]
Using the chain rule:
\[ \frac{-x}{\sqrt{1 - x^2}} + \frac{-y}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx} = a(1 - \frac{dy}{dx}) \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Rearranging gives:
\[ \frac{-y}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx} + a \frac{dy}{dx} = a - \frac{x}{\sqrt{1 - x^2}} \]
Factoring out \( \frac{dy}{dx} \):
\[ \left(a - \frac{y}{\sqrt{1 - y^2}}\right) \frac{dy}{dx} = a - \frac{x}{\sqrt{1 - x^2}} \]
Thus,
\[ \frac{dy}{dx} = \frac{a - \frac{x}{\sqrt{1 - x^2}}}{a - \frac{y}{\sqrt{1 - y^2}}} \]

### Final Result
This gives us the final expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} \]