Differentiation Question 382
Question: If $ \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y) $ , then $ \frac{dy}{dx}= $
[MNR 1983; ISM Dhanbad 1987; RPET 1991]
Options:
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}}} $
Show Answer
Answer:
Correct Answer: B
Solution:
Putting $ x=\sin \theta $ and $ y=\sin \varphi $
$ \cos \theta +\cos \varphi =a(\sin \theta -\sin \varphi ) $
Therefore $ 2\cos \frac{\theta +\varphi }{2}\cos \frac{\theta -\varphi }{2}=a{ 2\cos \frac{\theta +\varphi }{2}\sin \frac{\theta -\varphi }{2} } $
Therefore $ \frac{\theta -\varphi }{2}={{\cot }^{-1}}a\Rightarrow \theta -\varphi =2{{\cot }^{-1}}a $
Therefore $ {{\sin }^{-1}}x-{{\sin }^{-1}}y=2{{\cot }^{-1}}a $
Therefore $ \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}}} $ .
BETA
