# Choose the correct alternative in the following:

Question:

Choose the correct alternative in the following:

If $\sqrt{1-x^{6}}+\sqrt{1-y^{6}}=a^{3}\left(x^{3}-y^{3}\right)$, then $\frac{d y}{d x}$ is equal to

A. $\frac{\mathrm{x}^{2}}{\mathrm{y}^{2}} \sqrt{\frac{1-\mathrm{y}^{6}}{1-\mathrm{x}^{6}}}$

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

Solution:

$\sqrt{1-x^{6}}+\sqrt{1-y^{6}}=a^{3}\left(x^{3}-y^{3}\right)$

Let $x^{3}=\cos p$ and $y^{3}=\cos q$

$\cos ^{-1} x^{3}=p$ and $\cos ^{-1} y^{3}=q \cdots$ (1)

$\Rightarrow \sqrt{1-\cos ^{2} p}+\sqrt{1-\cos ^{2} q}=a^{3}(\cos p-\cos q)$

$\Rightarrow \sin p+\sin q=a(\cos p-\cos q)$

$\Rightarrow 2 \sin \left(\frac{p+q}{2}\right) \cdot \cos \left(\frac{p-q}{2}\right)=-2 a^{3} \sin \left(\frac{p-q}{2}\right) \cdot \sin \left(\frac{p+q}{2}\right)$

Comparing L.H.S and R.H.S we get,

$\Rightarrow \cos \left(\frac{p-q}{2}\right)=-a^{3} \sin \left(\frac{p-q}{2}\right)$

$\Rightarrow \frac{\sin \left(\frac{p-q}{2}\right)}{\cos \left(\frac{p-q}{2}\right)}=-\frac{1}{a^{3}}$

$\Rightarrow \tan \left(\frac{p-q}{2}\right)=-\frac{1}{a^{3}}$

$\Rightarrow \frac{p-q}{2}=\tan ^{-1}\left(-\frac{1}{a^{3}}\right)$

$\Rightarrow p-q=2 \cdot \tan ^{-1}\left(-\frac{1}{a^{3}}\right)$

Substituting value of $p$ and $q$ from (1)

$\Rightarrow \cos ^{-1}\left(\mathrm{x}^{3}\right)-\cos ^{-1}\left(\mathrm{y}^{3}\right)=2 \cdot \tan ^{-1}\left(-\frac{1}{\mathrm{a}^{3}}\right)$

Differentiating w.r.t $x$ we get,

$\Rightarrow-\frac{3 x^{2}}{\sqrt{1-x^{6}}}-\left(-\frac{3 y^{2}}{\sqrt{1-y^{6}}}\right) \cdot \frac{d y}{d x}=0$

$\Rightarrow\left(\frac{3 y^{2}}{\sqrt{1-y^{6}}}\right) \cdot \frac{d y}{d x}=\frac{3 x^{2}}{\sqrt{1-x^{6}}}$

Comparing L.H.S and R.H.S we get

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}^{2}}{\mathrm{y}^{2}} \sqrt{\frac{1-\mathrm{y}^{6}}{1-\mathrm{x}^{6}}}$