A) \[\frac{{{x}^{2}}}{{{y}^{2}}}\]
B) \[\frac{{{y}^{2}}}{{{x}^{2}}}\]
C) \[\frac{{{x}^{2}}+{{y}^{2}}}{{{y}^{2}}}\]
D) \[\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}}\]
Correct Answer: D
Solution :
Let \[x=\sqrt{{{a}^{{{\sin }^{-1}}}}t}\] \[\Rightarrow \]\[{{x}^{2}}={{a}^{{{\sin }^{-1}}t}}\Rightarrow 2\log \,x={{\sin }^{-1}}t.\log a\] \[\Rightarrow \]\[\frac{2}{x}=\frac{\log a}{\sqrt{1-{{t}^{2}}}}.\frac{dt}{dx}\] \[\Rightarrow \]\[\frac{2\sqrt{1-{{t}^{2}}}}{x\log \,a}=\frac{dt}{dx}\] ?(1) Now, let \[y=\sqrt{{{a}^{{{\cos }^{-1}}t}}}\] \[\Rightarrow \]\[2\log \,y={{\cos }^{-1}}t.\log a\] \[\Rightarrow \]\[\frac{2}{y}.\frac{dy}{dx}=\frac{-\log }{\sqrt{1-{{t}^{2}}}}.\frac{dt}{dx}\] \[\Rightarrow \]\[\frac{2}{y}.\frac{dy}{dx}=\frac{-\log a}{\sqrt{1-{{t}^{2}}}}\times \frac{2\sqrt{1-{{t}^{2}}}}{x\log a}\](from (1)) \[\Rightarrow \]\[\frac{dy}{dx}=-\frac{y}{x}\] Hence , \[1+{{\left( \frac{dy}{dx} \right)}^{2}}=1+{{\left( \frac{-y}{x} \right)}^{2}}=\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}}\]
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