Question

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

Hint:

Apply implicit differentiation systematically and simplify carefully.

Answer 1

Differentiate the given equation implicitly with respect to \( x \): \[ \frac{d}{dx} \left( \sqrt{1 - x^2} + \sqrt{1 - y^2} \right) = \frac{d}{dx} \left( a(x - y) \right). \] Applying the chain rule: \[ \frac{-x}{\sqrt{1 - x^2}} + \frac{-y}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx} = a(1 - \frac{dy}{dx}). \] Rearrange terms: \[ \frac{-x}{\sqrt{1 - x^2}} - a = \frac{dy}{dx} \left( a - \frac{y}{\sqrt{1 - y^2}} \right). \] Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{\frac{-x}{\sqrt{1 - x^2}} - a}{a - \frac{y}{\sqrt{1 - y^2}}}. \] For \( a = 0 \), simplify further to: \[ \frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}}. \] Conclusion: The result is proved.