Q. The two curves $x^{3} - 3xy^{2} + 2 = 0$ and $3x^{2}y - y^{3} - 2 = 0$
A) Cut at right angel
B) Touch each other
C) Cut at an angle $\frac{\pi}{3}$
D) Cot at an angle $\frac{\pi}{4}$
Explanation
(a)
On differentiating the given curves respectively
$\therefore\ \ \ \ \ \ \left( \frac{dy}{dx} \right)_{C_{1}} = \frac{x^{2} - y^{2}}{2xy} = m_{1}$ (say)
And $\left( \frac{dy}{dx} \right)_{C_{2}} = \frac{- 2xy}{x^{2} - y^{2}} = m_{2}$ (say)
$m_{1} \times m_{2} = - 1$
Hence, the two curves cut at right angles
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.