Question

If $\sqrt{(1-x^6)}+\sqrt{(1-y^6)}=a(x^3-y^3)$ and $\frac{dy}{dx}=f(x,y)\sqrt{\left(\frac{1-y^6}{1-x^6}\right)},$ then

• A. $f(x,y)=\frac{y}{x}$
• B. $f(x,y)=\frac{y^2}{x^2}$
• C. $f(x,y)=\frac{2y^2}{x^2}$
• D. $f(x,y)=\frac{x^2}{y^2}$

Answer 1

The correct option is D $f(x,y)=\frac{x^2}{y^2}$

Given : $\sqrt{(1-x^6)}+\sqrt{(1-y^6)}=a(x^3-y^3)$
Let, $x^3=\cos p$ and $y^3=\cos q$
$\Rightarrow \sqrt{1-{\cos }^{2}p}+\sqrt{1-{\cos }^{2}q}=a(\cos p-\cos q)\Rightarrow \sin p+\sin q=a(\cos p-\cos q)\Rightarrow 2\sin \left(\frac{p+q}{2}\right)\cos \left(\frac{p-q}{2}\right)=-2a\sin \left(\frac{p-q}{2}\right)\sin \left(\frac{p+q}{2}\right)\Rightarrow \tan \left(\frac{p-q}{2}\right)=-\frac{1}{a}\Rightarrow p-q=2{\tan }^{-1}(-\frac{1}{a})\Rightarrow {\cos }^{-1}{x}^3-{\cos }^{-1}{y}^3=2{\tan }^{-1}(-\frac{1}{a})$
Differentiating w.r.t. $x$ both side, we have
$-\frac{3x^2}{\sqrt{1-x^6}}+\frac{3y^2}{\sqrt{1-y^6}}\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{x^2}{y^2}\sqrt{\frac{1-y^6}{1-x^6}}$
Hence, $f(x,y)=\frac{x^2}{y^2}$