The two curves x3−3xy2+2=0 and 3x2y−y3−2=0 intersect at an angle of whose degree measure is

For the curve $x^{3} - 3 x y^{2} + 2 = 0$ , the gradient is $\frac{d y}{d x} = - \frac{3 x^{2} - 3 y^{2}}{- 6 x y} = \frac{x^{2} - y^{2}}{2 x y}$

For the curve $3 x^{2} y - y^{3} - 2 = 0$ , the gradient is $\frac{d y}{d x} = - \frac{6 x y}{3 x^{2} - 3 y^{2}} = - \frac{2 x y}{x^{2} - y^{2}}$

Observing the above two gradients, the product of slopes is -1

hence, the angle between the curves is $90^{0}$

The two curves $x^{3} - 3 x y^{2} + 2 = 0$ and $3 x^{2} y - y^{3} - 2 = 0$
intersect at an angle of whose degree measure is