Question

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

• A. touch each other
• B. cut each other at right angle
• C. cut at an angle $\frac{\pi }{3}$
• D. cut at an angle $\frac{\pi }{4}$

Answer 1

The correct option is B cut each other at right angle

$x^3-3x{y}^2+2=0$
$\Rightarrow \frac{dy}{dx}=\frac{x^2-y^2}{2xy}=m_1\text{(say)}$

$3x^{2}y-y^3=2$
$\Rightarrow \frac{dy}{dx}=\frac{-2xy}{x^2-y^2}=m_2\text{(say)}$
Product of the slopes at any point $(x,y)=m_1\times m_2=-1$
Therefore, the curves are orthogonal and cut each other at right angle.