Question

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

• A. $1$
• B. $\frac{x}{y}$
• C. $\frac{x^{n-1}}{y^{n-1}}$
• D. $Noneofthese$

Answer 1

Answer: (C)

$\frac{x^{n-1}}{y^{n-1}}$

$\sqrt{1-x^{2n}}+\sqrt{1-y^{2n}}=a\left(x^n-y^n\right),$

$\frac{1}{2\sqrt{1-x^{2n}}}\times (-2n{x}^{2n-1})+\frac{1}{2\sqrt{1-y^{2n}}}\times (-2n{y}^{2n-1})\frac{dy}{dx}=an{x}^{n-1}n{y}^{n-1}\frac{dy}{dx}$

$\frac{-x^{2n-1}}{\sqrt{1-x^{2n}}}-\frac{y^{2n-1}}{\sqrt{1-y^{2n}}}dfracdydx=a{x}^{n-1}-a{y}^{n-1}\frac{dy}{dx}$

$\Rightarrow a{y}^{n-1}\frac{dy}{dx}-a{y}^{n-1}\frac{dy}{dx}$

$\Rightarrow \frac{\sqrt{1-x^{2n}}+\sqrt{1-y^{2n}}}{x^n-y^n}.y^{n-1}\frac{dy}{dx}-\frac{y^{2n-1}}{\sqrt{1-y^{2n}}}\frac{dy}{dx}=$

$\frac{\sqrt{1-x^{2n}}+\sqrt{1-y^{2n}}}{x^n-y^n}.x^{n-1}+\frac{x^{2n-1}}{\sqrt{1-x^{2n}}}$

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

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