Question

If $\sqrt{1-x^6}+\sqrt{1-y^6}=a\left(x^3-y^3\right),$ then $y^2\frac{dy}{dx}=$

• A. $\sqrt{\frac{1-y^6}{1-x^6}}$
• B. $x\sqrt{\frac{1-y^6}{1-x^6}}$
• C. $x^2\sqrt{\frac{1-y^6}{1-x^6}}$
• D. $\frac{1}{x^2}\sqrt{\frac{1-y^6}{1-x^6}}$

Answer 1

Answer: (C)

Given equation is
$\sqrt{1-x^6}+\sqrt{1-y^6}=a\left(x^3-y^3\right)$
$\Rightarrow \frac{\sqrt{1-x^6}+\sqrt{1-y^6}}{x^3-y^3}=a$
On differentiating both sides w.r.t., $x$, we get
$\frac{\left[\left(x^3-y^3\right)\left(\frac{-6x^5}{2\sqrt{1-x^6}}-\frac{6y^5}{2\sqrt{1-y^6}}\frac{dy}{dx}\right)-\left(\sqrt{1-x^6}+\sqrt{1-y^6}\right)\left(3x^2-3y^2\frac{dy}{dx}\right)\right]}{{\left(x^3-y^3\right)}^2}=0$
$\Rightarrow \left(y^2\left(\sqrt{1-x^6}+\sqrt{1-y^6}\right)-\frac{y^5\left(x^3-y^3\right)}{\sqrt{1-y^6}}\right)\frac{dy}{dx}$
$=x^2\left(\sqrt{1-x^6}+\sqrt{1-y^6}\right)+\frac{x^5}{\sqrt{1-x^6}}\left(x^3-y^3\right)$
$\Rightarrow y^2\frac{dy}{dx}\left[\frac{\sqrt{1-x^6}+\sqrt{1-y^6}+\left(1-y^6\right)-y^3{x}^3+y^6}{\sqrt{1-y^6}}\right]$
$=x^2\left[\frac{\left(1-x^6\right)+\sqrt{1-y^6}\sqrt{1-x^6}+x^6-x^3{y}^3}{\sqrt{1-x^6}}\right]$
$\Rightarrow y^2\frac{dy}{dx}\left[\frac{\sqrt{1-x^6}\sqrt{1-y^6}+1-x^3{y}^3}{\sqrt{1-y^6}}\right]$
$=x^2\left[\frac{1+\sqrt{1-y^6}\sqrt{1-x^6}-x^3{y}^3}{\sqrt{1-x^6}}\right]$
$\Rightarrow y^2\frac{dy}{dx}=x^2\sqrt{\frac{1-y^6}{1-x^6}}$