A) cut at right angles
B) touch each other
C) cut at an angle\[\frac{\pi }{3}\]
D) cut at an angle\[\frac{\pi }{4}\]
Correct Answer: A
Solution :
Given curves are \[{{x}^{3}}-3x{{y}^{2}}+2=0\] and \[3{{x}^{2}}y-{{y}^{3}}-2=0\] On differentiating w.r.t. x, we get \[{{\left( \frac{dy}{dx} \right)}_{{{c}_{1}}}}=\frac{{{x}^{2}}-{{y}^{2}}}{2xy}\] and \[{{\left( \frac{dy}{dx} \right)}_{{{c}_{2}}}}=\frac{-2xy}{{{x}^{2}}-{{y}^{2}}}\] Now, \[{{\left( \frac{dy}{dx} \right)}_{{{c}_{1}}}}\times {{\left( \frac{dy}{dx} \right)}_{{{c}_{2}}}}=-1\] Hence, the two curves cut at right angles.
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