A) cut at right angle
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 equations of two curves are \[{{x}^{3}}-3x{{y}^{2}}+2=0\] ... (i) and \[3{{x}^{2}}y\,-{{y}^{2}}-2=0\] ... (ii) On differentiating Eqs. (i) and (ii) 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. \[{{\left( \frac{dy}{dx} \right)}_{{{C}_{1}}}}\times {{\left( \frac{dy}{dx} \right)}_{{{C}_{2}}}}=\left( \frac{{{x}^{2}}-{{y}^{2}}}{2xy} \right)\left( \frac{-2xy}{{{x}^{2}}-{{y}^{2}}} \right)\] \[=-1\] Hence, the two curves cut at right angle.
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