Q110 / 180 · KCET 2015 · 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 at right angle
  • C
    cut at angle \( \frac{\Pi}{3} \)
  • D
    cut at angle \( \frac{\Pi}{4} \)

Solution

Given Curves, $x^{3}-3 x y^{2}+2=0 \rightarrow(1)$ $3 x^{2} y-y^{3}=2 \rightarrow(2)$ Differentiating Eqs. (1) and (2) with respect to x, we have $3 x^{2}-3\left(y^{2}+x(2 y y)\right)=0$ $\Rightarrow x^{2}=y^{2}+2 x y y$ $\Rightarrow y=\frac{x^{2}-y^{2}}{2 x y}=m_{1}$ (let) $\rightarrow(3)$ $3\left(x^{2} y+2 x y\right)-3 y^{2} y=0$ $\Rightarrow x^{2} y+2 x y-y^{2} y=0$ $\Rightarrow y=-\frac{2 x y}{x^{2}-y^{2}}=m_{2}$ (let) $\rightarrow$ (4) From Eqs. (3) and (4), we have $m 1 \cdot m_{2}=-1$ Hence, these curves cut at $90^{\circ}$

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

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