A) touch each other
B) cut a an angle \[\pi /4\]
C) cut at an angle \[\pi /3\]
D) cut at an angle \[\pi /2\]
Correct Answer: D
Solution :
Given curves are \[{{x}^{3}}-3x{{y}^{2}}+2=0\] ?..(i) and \[3{{x}^{2}}y-{{y}^{3}}-2=0\] ?..(ii) On differentiating Eq. (i) w. r. t .x, we get \[3{{x}^{2}}-3x2y\frac{dy}{dx}-3{{y}^{2}}+0=0\] \[\Rightarrow \] \[6xy\frac{dy}{dx}=3({{x}^{2}}-{{y}^{2}})\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{{{x}^{2}}-{{y}^{2}}}{2xy}={{m}_{1}}\] On differentiating Eq. (ii), w. r. t .x, we get \[6xy+3{{x}^{2}}\frac{dy}{dx}-3{{y}^{2}}\frac{dy}{dx}+0=0\] \[\Rightarrow \] \[6xy+3({{x}^{2}}-{{y}^{2}})\frac{dy}{dx}=0\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{-6xy}{3({{x}^{2}}-{{y}^{2}})}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{-2xy}{{{x}^{2}}-{{y}^{2}}}={{m}_{2}}\] Now, \[{{m}_{1}}\times {{m}_{2}}=\frac{{{x}^{2}}-{{y}^{2}}}{2xy}\times \frac{-2xy}{{{x}^{2}}-{{y}^{2}}}=-1\] \[\therefore \] Two curves cut at an angle of \[\frac{\pi }{2}\].
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