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
(a) $\frac{x^2}{y^2}\sqrt{\frac{1-y^6}{1-x^6}}$

(b) $\frac{y^2}{x^2}\sqrt{\frac{1-y^6}{1+x^6}}$

(c) $\frac{x^2}{y^2}\sqrt{\frac{1-x^6}{1-y^6}}$

(d) none of these

Answer 1

(a) $\frac{x^2}{y^2}\sqrt{\frac{1-y^6}{1-x^6}}$

$$\begin{gathered} \mathrm{We}\mathrm{have},\sqrt{1-x^6}+\sqrt{1-y^6}=a\left(x^3-y^3\right) \\ \mathrm{Putting}{x}^3=\sin A\mathrm{and}{y}^3=\sin B \\ \Rightarrow \sqrt{1-{\sin }^{2}A}+\sqrt{1-{\sin }^{2}B}=a\left(\sin A-\sin B\right) \\ \Rightarrow \cos A+\cos B=a\left(\sin A-\sin B\right) \\ \Rightarrow 2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)=2a\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right) \\ \Rightarrow \cot \left(\frac{A-B}{2}\right)=a \\ \Rightarrow \frac{A-B}{2}={\cot }^{-1}\left(a\right) \\ \Rightarrow A-B=2{\cot }^{-1}\left(a\right) \\ \Rightarrow {\sin }^{-1}{x}^3-{\sin }^{-1}{y}^3=2{\cot }^{-1}\left(a\right) \end{gathered}$$

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