A) \[\frac{dy}{dx}=\frac{{{x}^{2}}+{{C}^{2}}}{{{x}^{2}}-{{C}^{2}}}\]
B) \[\frac{dy}{dx}=\frac{{{x}^{2}}-{{C}^{2}}}{{{x}^{2}}+{{C}^{2}}}\]
C) \[\frac{dy}{dx}=-\frac{{{C}^{2}}}{{{x}^{2}}}\]
D) None of these
Correct Answer: B
Solution :
[b] Let the slope of tangent of required family be \[\frac{dy}{dx}={{m}_{1}}\] Also\[y=\frac{{{C}^{2}}}{x}\]; therefore, \[\frac{dy}{dx}=-\frac{{{C}^{2}}}{{{x}^{2}}}={{m}^{2}}\] (say). By the given condition, we have \[\tan \frac{\pi }{4}\] \[=\frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}\Rightarrow 1+{{m}_{1}}{{m}_{2}}={{m}_{1}}-{{m}_{2}}\] \[\Rightarrow \frac{dy}{dx}+\frac{{{C}^{2}}}{{{x}^{2}}}=1-\frac{{{C}^{2}}}{{{x}^{2}}}\frac{dy}{dx}\Rightarrow \frac{dy}{dx}\left( 1+\frac{{{C}^{2}}}{{{x}^{2}}} \right)\] \[=1-\frac{{{C}^{2}}}{{{x}^{2}}}\Rightarrow \frac{dy}{dx}=\frac{{{x}^{2}}-{{C}^{2}}}{{{x}^{2}}+{{C}^{2}}}\]
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