Q91 / 180 · KCET 2016 · Mathematics

The two curves \( x^{3}-3 x y^{2}+2=0 \) and \( 3 x^{2} y-y^{3}=2 \)

  • A
    Touch each other
  • Correct answer
    B
    Cut each other at right angle
  • C
    Cut at an angle \( \frac{\Pi}{3} \)
  • D
    Cut at an angle \( \frac{\Pi}{4} \)

Solution

Given curves, $\rightarrow$ $x^{3}-3 x y^{2}+2=0 \rightarrow(1)$ $3 x^{2} y-y^{3}=2 \rightarrow(2)$ Differentiating Eqs. (1) and (2) both the sides with respect to $x$, we get $3 x^{2}-3\left(y^{2}+x\left(2 y y^{\prime}\right)\right)=0$ $\Rightarrow x^{2}=y^{2}+2 x y y^{\prime}$ $\Rightarrow y^{\prime}=\frac{x^{2}-y^{2}}{2 x y}=m_{1}($ Let $) \rightarrow(3)$ $3\left(x^{2} y^{\prime}+2 x y\right) 3 y^{2} y=0$ $\Rightarrow x^{2} y^{\prime}+2 x y-y^{2} y^{\prime}=0$ $\Rightarrow y^{\prime}=\frac{-2 x y}{x^{2}-y^{2}}=m_{2}($ Let $) \rightarrow(4)$ From Eqs. (3) and (4), we get $m_{1} m_{2}=-1$ Hence, the two curves cut at an angle $\frac{\Pi}{2}$ i.e. right angle

Hence, the correct answer is option: 'B'.

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