Question

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

Hint:

Apply the chain rule carefully for derivatives of composite functions like $\sqrt{1 - x^2}$ and $\sqrt{1 - y^2}$. Rearrange terms systematically to isolate $\frac{dy}{dx}$.

Answer 1

The given equation is: $$\sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y).$$ 

Differentiate both sides with respect to $x$: $$\frac{d}{dx}\left(\sqrt{1 - x^2}\right) + \frac{d}{dx}\left(\sqrt{1 - y^2}\right) = \frac{d}{dx}[a(x - y)].$$ Using the chain rule: $$\frac{-x}{\sqrt{1 - x^2}} + \frac{-y}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx} = a(1 - \frac{dy}{dx}).$$ 

Rearrange the terms: $$\frac{-y}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx} + a\frac{dy}{dx} = a - \frac{x}{\sqrt{1 - x^2}}.$$ 

Factorize $\frac{dy}{dx}$ on the left-hand side: $$\frac{dy}{dx}\left(a - \frac{y}{\sqrt{1 - y^2}}\right) = a - \frac{x}{\sqrt{1 - x^2}}.$$ 

Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{a - \frac{x}{\sqrt{1 - x^2}}}{a - \frac{y}{\sqrt{1 - y^2}}}.$$ 

For the given condition $a = 1$, this simplifies to: $$\frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}}.$$ Hence, it is proved that: $$\frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}}.$$