Question

If \( \sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y) \), then prove that \( \frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} \).

Hint:

When differentiating equations involving square roots, apply the chain rule carefully, and remember that \( \frac{d}{dx} \sqrt{1 - x^2} = \frac{-x}{\sqrt{1 - x^2}} \).

Answer 1

The provided equation is: \[ \sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y) \] Differentiating both sides with respect to \( x \): The derivative of the left-hand side is: \[ \frac{d}{dx} \left( \sqrt{1 - x^2} \right) = \frac{-x}{\sqrt{1 - x^2}} \] and using the chain rule for the term involving \( y \): \[ \frac{d}{dx} \left( \sqrt{1 - y^2} \right) = \frac{-y \frac{dy}{dx}}{\sqrt{1 - y^2}} \] The derivative of the right-hand side is: \[ \frac{d}{dx} \left( a(x - y) \right) = a \left( 1 - \frac{dy}{dx} \right) \] Combining these, the differentiated equation is: \[ \frac{-x}{\sqrt{1 - x^2}} + \frac{-y \frac{dy}{dx}}{\sqrt{1 - y^2}} = a \left( 1 - \frac{dy}{dx} \right) \] Rearranging to solve for \( \frac{dy}{dx} \): \[ \frac{-y \frac{dy}{dx}}{\sqrt{1 - y^2}} + a \frac{dy}{dx} = \frac{x}{\sqrt{1 - x^2}} + a \] Factoring out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \left( a - \frac{y}{\sqrt{1 - y^2}} \right) = \frac{x}{\sqrt{1 - x^2}} + a \] Solving for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{\frac{x}{\sqrt{1 - x^2}} + a}{a - \frac{y}{\sqrt{1 - y^2}}} \] This simplifies to: \[ \frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} \] This completes the proof.