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 $f(x,y)=\frac{x^2}{y^2}$

$Put{x}^3=sin\theta ,y^3=sin\varphi ,$
$thencos\theta +cos\varphi =a(sin\theta -sin\varphi )$
$\Rightarrow 2cos\left(\frac{\theta +\varphi }{2}\right)cos\left(\frac{\theta -\varphi }{2}\right)$ $=2acos\left(\frac{\theta +\varphi }{2}\right)sin\left(\frac{\theta -\varphi }{2}\right)$
$\Rightarrow cot\left(\frac{\theta -\varphi }{2}\right)$ = a
$\Rightarrow \left(\frac{\theta -\varphi }{2}\right)=co{t}^{-1}a$
$\Rightarrow si{n}^{-1}{x}^3-si{n}^{-1}{y}^3=2co{t}^{-1}a$
$\therefore \frac{3x^2}{\sqrt{(1-x^6)}}-\frac{3y^2}{\sqrt{(1-y^6)}}\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{x^2}{y^2}\sqrt{\left(\frac{1-y^6}{1-x^6}\right)}$
$\therefore f(x,y)=\frac{x^2}{y^2}$