Question

If \[y \sqrt{1 - x^2} + x \sqrt{1 - y^2} = 1\] ,prove that \[\frac{dy}{dx} = - \sqrt{\frac{1 - y^2}{1 - x^2}}\] ?

Answer 1

\[\text{We have }, y\sqrt{1 - x^2} + x\sqrt{1 - y^2} = 1\]

\[\text{Let x  }= \sin A , y = \sin B\]

\[ \Rightarrow \sin B\sqrt{1 - \sin^2 A} + \sin A\sqrt{1 - \sin^2 B} = 1\]

\[ \Rightarrow \sin B\cos A + \sin A\cos B = 1 \left[ \because \sin\left( x + y \right) = \sin x \cos y + \cos x\sin y \right]\]

\[ \Rightarrow \sin\left( A + B \right) = 1\]

\[ \Rightarrow A + B = \sin^{- 1} \left( 1 \right)\]

\[ \Rightarrow \sin^{- 1} x + \sin^{- 1} y = \frac{\pi}{2} \left[ \because x = \sin A, y = \sin B \right]\]

Differentiate with respect to x,

\[\Rightarrow \frac{d}{dx}\left( \sin^{- 1} x \right) + \frac{d}{dx}\left( \sin^{- 1} y \right) = \frac{d}{dx}\left( \frac{\pi}{2} \right)\]

\[ \Rightarrow \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - y^2}}\frac{d y}{d x} = 0\]

\[ \Rightarrow \frac{d y}{d x} = - \sqrt{\frac{1 - y^2}{1 - x^2}}\]