A) \[\frac{y}{x}\]
B) \[\frac{x}{y}\]
C) \[\frac{-y}{x}\]
D) \[\frac{-x}{y}\]
Correct Answer: C
Solution :
Given, \[x=\sqrt{{{a}^{{{\sin }^{-1}}t}}}\] ...(i) and \[y=\sqrt{{{a}^{{{\cos }^{-1}}t}}}\] ...(ii) On multiplying Eqs. (i) and (ii), we get \[xy=\sqrt{{{a}^{{{\sin }^{-1}}t}}}\times \sqrt{{{a}^{{{\cos }^{-1}}t}}}\] \[\Rightarrow \] \[xy=\sqrt{{{a}^{{{\sin }^{-1}}t}}.{{a}^{{{\cos }^{-1}}t}}}\] \[=\sqrt{{{a}^{{{(}^{{{\sin }^{-1}}t+}}^{{{\cos }^{-1}}t})}}}\] \[\left( \because \,{{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2} \right)\] \[\Rightarrow \] \[xy=\sqrt{{{a}^{\pi /12}}}\] On differentiating w.r.t. x, we get \[x\frac{dy}{dx}+y=0\Rightarrow x\frac{dy}{dx}=-y\Rightarrow \frac{dy}{dx}=\frac{-y}{x}\] \[\therefore \] \[\frac{d}{dx}(dconstant)=0\]
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