Question

If  \[\sqrt{1 - x^6} + \sqrt{1 - y^6} = a^3 \left( x^3 - y^3 \right)\] then \[\frac{dy}{dx}\] is equal to ____________ .

Options

• \[\frac{x^2}{y^2} \sqrt{\frac{1 - y^6}{1 - x^6}}\]

• \[\frac{y^2}{x^2}\sqrt{\frac{1 - y^6}{1 + x^6}}\]

• \[\frac{x^2}{y^2}\sqrt{\frac{1 - x^6}{1 - y^6}}\]

• none of these

Answer 1

\[\frac{x^2}{y^2} \sqrt{\frac{1 - y^6}{1 - x^6}}\]

 

\[\text { We have }, \sqrt{1 - x^6} + \sqrt{1 - y^6} = a\left( x^3 - y^3 \right)\]
\[\text { Putting } x^3 = \sin A \text { and }y^3 = \sin B\]
\[ \Rightarrow \sqrt{1 - \sin^2 A} + \sqrt{1 - \sin^2 B} = a\left( \sin A - \sin B \right)\]
\[ \Rightarrow \cos A + \cos B = a\left( \sin A - \sin B \right)\]
\[ \Rightarrow 2\cos\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) = 2a \sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right)\]
\[ \Rightarrow \cot\left( \frac{A - B}{2} \right) = a\]
\[ \Rightarrow \frac{A - B}{2} = \cot^{- 1} \left( a \right)\]
\[ \Rightarrow A - B = 2 \cot^{- 1} \left( a \right)\]
\[ \Rightarrow \sin^{- 1} x^3 - \sin^{- 1} y^3 = 2 \cot^{- 1} \left( a \right)\]

\[\Rightarrow \frac{1}{\sqrt{1 - x^6}} \times \frac{d}{dx}\left( x^3 \right) - \frac{1}{\sqrt{1 - y^6}} \times \frac{d}{dx}\left( y^3 \right) = 0\]
\[ \Rightarrow \frac{1}{\sqrt{1 - x^6}} \times 3 x^2 - \frac{1}{\sqrt{1 - y^6}} \times 3 y^2 \times \frac{dy}{dx} = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x^2}{y^2}\sqrt{\frac{1 - y^6}{1 - x^6}}\]