A) \[y\sqrt{1-{{y}^{2}}}+x\sqrt{1-{{x}^{2}}}=a\]
B) \[x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}}=a\]
C) \[x\sqrt{1-{{y}^{2}}}-y\sqrt{1-{{x}^{2}}}=a\]
D) \[y\sqrt{1-{{y}^{2}}}-x\sqrt{1-{{x}^{2}}}=a\]
Correct Answer: B
Solution :
[b]: We have,\[\frac{dy}{dx}+\sqrt{\frac{1-{{y}^{2}}}{1-{{x}^{2}}}}=0\] \[\Rightarrow \]\[\frac{dy}{dx}=-\sqrt{\frac{1-{{y}^{2}}}{1-{{x}^{2}}}}\]\[\Rightarrow \]\[\frac{dy}{\sqrt{1-{{y}^{2}}}}=-\frac{dx}{\sqrt{1-{{x}^{2}}}}\] \[\Rightarrow \]\[\int_{{}}^{{}}{\frac{dy}{\sqrt{1-{{y}^{2}}}}}=-\int_{{}}^{{}}{\frac{dx}{\sqrt{1-{{x}^{2}}}}}\] \[\Rightarrow \]\[{{\sin }^{-1}}y+{{\sin }^{-1}}x={{\sin }^{-1}}a\] \[\Rightarrow \]\[x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}}=a\]
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