An angle at which the two curves $x^{3} - 3 x y^{2} = 4$ and $3 x^{2} y - y^{3} = 4$ intersect is

\frac{π}{2}

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\frac{π}{4}

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\frac{π}{3}

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Solution

The correct option is A $\frac{π}{2}$
$x y^{2} = x^{3} - 4$
$y^{2} = \frac{x^{3} - 4}{x}$
$y = (\frac{x^{3} - 4}{x})$
$3 x^{2} - 3 y^{2} - 6 x y y^{1} = 0$
$y^{'} = \frac{x^{2} - y^{2}}{2 x y} = m_{1}$
$3 x^{2} y^{1} + 6 x y - 3 y^{2} y^{1} = 0$
$y^{'} = \frac{- 2 x y}{x^{2} - y^{2}} = m_{2}$
$m_{1} \cdot m_{2} = - 1$
$∴$ Angle b/w the curve is $^{π} /_{2}$

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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

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

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

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} + 2 = 0

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

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