Question

The angle between the two curves $x^3–3x{y}^2=4\text{and}3x^{2}y–y^3=4$ is

• A. $\pi /2$
• B. $\pi /4$
• C. $\pi /3$
• D. $\pi /6$

Answer 1

The correct option is A

$\pi /2$

$m_1=-\left(\frac{3x^2-3y^2}{-6xy}\right)$ and

$m_2=-\left(\frac{6xy}{3x^2-3y^2}\right)$

$m_1{m}_2=-1\therefore \theta =\frac{\pi }{2}$