Differentiation Question 50
Question: If $ \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y) $ , then $ \frac{dy}{dx} $ =
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}}} $
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Answer:
Correct Answer: B
Solution:
[b] Putting $ x=\sin \theta $ and $ y=\sin \phi $
$ \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}}} $
BETA
