Question

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

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

Answer 1

The correct option is A $\frac{x^{n-1}}{y^{n-1}}$

Put $x^n=sin\theta$ and $yn=sin\varphi$
then, $(cos\theta +cos\varphi )=a(sin\theta -sin\varphi )$
$\Rightarrow 2cos\left(\frac{\theta +\varphi }{2}\right)cos\left(\frac{\theta -\varphi }{2}\right)=2acos\left(\frac{\theta +\varphi }{2}\right)sin\left(\frac{\theta -\varphi }{2}\right)\Rightarrow cot\left(\frac{\theta -\varphi }{2}\right)=a\Rightarrow \left(\frac{\theta -\varphi }{2}\right)=co{t}^{-1}a\Rightarrow \theta -\varphi =2co{t}^{-1}a$
or $si{n}^{-1}{x}^n-si{n}^{-1}{y}^n=2co{t}^{-1}a$
Differentiating both sides, we have
$\frac{n{x}^{n-1}}{\sqrt{1-x^{2}n}}-\frac{n{y}^{n-1}}{\sqrt{(1-y^{2n})}}\frac{dy}{dx}=0\therefore \sqrt{(\frac{1-x^{2}n}{1-y^{2n}}}\frac{dy}{dx}=\frac{x^n-1}{y^n-1}$