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?

\frac{π}{4}

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\frac{π}{3}

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\frac{π}{2}

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\frac{π}{6}

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Solution

The correct option is C $\frac{π}{2}$
correct answer is c)

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Similar questions

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

x^{3} - 3 x y^{2} + 2 = 0

3 x^{2} y - y^{3} = 2

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

x^{3} - 3 x y^{2} = 4

3 x^{2} y - y^{3} = 4

x^{3} - 3 x y^{2} + 2 = 0

3 x^{2} y - y^{3} - 2 = 0

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