Question

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

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

Answer 1

The correct option is A cut at right angles

Curves are $C_1=x^3-3x{y}^2+2=0$ and $C_2={-y}^3+3y{x}^2-2=0$

Suppose they intersect at $(m,n)$

So there will be two tangents to both curves at $(m,n)$

And their slopes are $a$ and $b$

$\frac{d\left(C_1\right)}{dx}={3m}^2-3(n^2+2mna)⟹a=\frac{m^2-n^2}{2mn}$

$\frac{d\left(C_2\right)}{dx}={-3n}^2+3(m^2+2mnb)⟹b=\frac{2mn}{n^2-m^2}$

So $a\cdot b=-1$

This means both curves are orthogonal at intersection.

So they cut at right angles.