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 $y = sin\,\theta$
$cos \,\theta + cos \,\phi = a(sin\, \theta - sin\,\phi)$
$\Rightarrow 2\,cos \frac{\theta + \phi}{2} cos \frac{\theta - \phi}{2} = a \{ 2\,cos \frac{\theta + \phi}{2} sin \frac{\theta - \phi}{2}\}$
$\Rightarrow \frac{\theta - \phi}{2} = cos^{-1} a$
$\Rightarrow \theta - \phi = 2\, cot^{-1} a$
$\Rightarrow sin^{-1} x - sin^{-1} y = 2\, cot^{-1} a$
$\Rightarrow \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - y^2}} \frac{dy}{dx} = 0$
$\Rightarrow \frac{dy}{dx} = \sqrt {\frac{ 1 - y^2}{1 - x^2}}$