Question

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

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

Answer 1

Answer: (B)

Putting $x=sin\theta$ and $y=sin\theta$
$cos\theta +cos\varphi =a(sin\theta -sin\varphi )$
$\Rightarrow 2cos\frac{\theta +\varphi }{2}cos\frac{\theta -\varphi }{2}=a\{2cos\frac{\theta +\varphi }{2}sin\frac{\theta -\varphi }{2}\}$
$\Rightarrow \frac{\theta -\varphi }{2}=co{s}^{-1}a$
$\Rightarrow \theta -\varphi =2co{t}^{-1}a$
$\Rightarrow si{n}^{-1}x-si{n}^{-1}y=2co{t}^{-1}a$
$\Rightarrow \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$