A) \[x{{y}_{1}}=n\sqrt{{{b}^{2}}-{{y}^{2}}}\]
B) \[x{{y}_{1}}+n\sqrt{{{b}^{2}}-{{y}^{2}}}=0\]
C) \[{{y}_{1}}=x\sqrt{{{b}^{2}}-{{y}^{2}}}\]
D) \[x{{y}_{1}}-\sqrt{{{b}^{2}}-{{y}^{2}}}=0\]
Correct Answer: B
Solution :
Given, \[{{\cos }^{-1}}\left( \frac{y}{b} \right)=n\log \left( \frac{x}{n} \right)\] \[\left( \because \,{{y}_{1}}=\frac{dy}{dx} \right)\] Differentiating w.r.t. 'x' \[-\frac{1}{\sqrt{1-{{(y/b)}^{2}}}}.\frac{{{y}_{1}}}{b}=n.\frac{1}{(x/n)}.\frac{1}{n}\] \[-\frac{b}{\sqrt{{{b}^{2}}-{{y}^{2}}}}.\frac{{{y}_{1}}}{b}=\frac{{{n}^{2}}}{x}.\frac{1}{n}\] \[\Rightarrow \] \[-\frac{{{y}_{1}}}{\sqrt{{{b}^{2}}-{{y}^{2}}}}=\frac{n}{x}\] \[\Rightarrow \] \[-x{{y}_{1}}=n\sqrt{{{b}^{2}}-{{y}^{2}}}\] \[\Rightarrow \] \[x{{y}_{1}}+n\sqrt{{{b}^{2}}-{{y}^{2}}}=0\]
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