Q118 / 200 · Manipal (MET) 2011 · Mathematics
The two curves $x^3-3 x y^2+2=0$ and $3 x^2 y-y^3=2$
- Correct answerAcut at right angles
- Btouch each other
- Ccut at an angle $\pi / 3$
- Dcut at an angle $\pi / 4$
Solution
Given curves are
$x^3-3 x y^2+2=0$..(i)
and $\quad 3 x^2 y-y^3-2=0$...(ii)
On differentiating Eqs. (i) and (ii), with respect to $x$, we get
$\left(\frac{d y}{d x}\right) C_1=\frac{x^2-y^2}{2 x y}$
$\begin{aligned}\left(\frac{d y}{d x}\right) C_2 & =\frac{2 x y}{x^2-y^2} \\ \left(\frac{d y}{d x}\right)_{C_1} \times\left(\frac{d y}{d x}\right)_{C_2} & =-1\end{aligned}$
$\begin{aligned}\left(\frac{d y}{d x}\right) C_2 & =\frac{2 x y}{x^2-y^2} \\ \left(\frac{d y}{d x}\right)_{C_1} \times\left(\frac{d y}{d x}\right)_{C_2} & =-1\end{aligned}$
Hence, the correct answer is option: 'A'.