If \[\sqrt{1-{{x}^{2}}}+\sqrt{1-{{y}^{2}}}=a(x-y)\], then \[\frac{dy}{dx}=\] [MNR 1983; ISM Dhanbad 1987; RPET 1991]

\[\sqrt{\frac{{{x}^{2}}-1}{1-{{y}^{2}}}}\]

\[\sqrt{\frac{1-{{x}^{2}}}{1-{{y}^{2}}}}\]

\[\sqrt{\frac{1-{{y}^{2}}}{1-{{x}^{2}}}}\]

\[\sqrt{\frac{{{y}^{2}}-1}{1-{{x}^{2}}}}\]

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

1.	If \[\sqrt{1-{{x}^{2}}}+\sqrt{1-{{y}^{2}}}=a(x-y)\], then \[\frac{dy}{dx}=\] [MNR 1983; ISM Dhanbad 1987; RPET 1991]
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» C. \[\sqrt{\frac{{{x}^{2}}-1}{1-{{y}^{2}}}}\]

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If \[\sqrt{1-{{x}^{2}}}+\sqrt{1-{{y}^{2}}}=a(x-y)\], then \[\frac{dy}{dx}=\] [MNR 1983; ISM Dhanbad 1987; RPET 1991]

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