Question

The two curves $x^3 - 3xy^2 + 2 = 0$ and $3x^2y - y^3 = 2$

• A. Touch each other
• B. Cut each other at right angle
• C. Cut at an angle $\pi / 3$
• D. Cut at an angle $\pi / 4$

Answer 1

Answer: (B)

Cut each other at right angle

Answer 2

We have, $x^{3}-3 x y^{2}+2=0$
$\Rightarrow 3 x^{2}-6 x y \frac{d y}{d x}-3 y^{2}=0$
$\Rightarrow \frac{d y}{d x}=\frac{3\left(x^{2}-y^{2}\right)}{6 x y}$
Now, $\left(\frac{d y}{d x}\right)_{(h, k)}=\frac{3\left(h^{2}-k^{2}\right)}{6 h k}=m_{1}$[say]
and $3 x^{2} y-y^{8}=2$
$\Rightarrow 3 x^{2} \frac{d y}{d x}+6 x y-3 y^{2} \frac{d y}{d x}=0$
$\Rightarrow \frac{d y}{d x}=\frac{-6 x y}{3\left(x^{2}-y^{2}\right)}$
Now, $\left(\frac{d y}{d x}\right)_{(h, h)}=\frac{-6 h k}{3\left(h^{2}-k^{2}\right)}=m_{2}$[say]
$\therefore m_{1} \cdot m_{2}=\frac{3\left(h^{2}-k^{2}\right)}{6 h k} \times \frac{-6 h k}{3\left(h^{2}-k^{2}\right)}=-1$
Hence, both the curves cut each other at right angle.