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Q.

An angle between the curves $x^{3} + y^{3} + x + 2 y = 0, x y + 2 x = y$ at the origin is

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a

$\frac{π}{3}$

b

$0$

c

$\frac{π}{4}$

d

$\frac{π}{2}$

answer is C.

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Detailed Solution

$\begin{aligned} x^{3} + y^{3} + x + 2 y = 0 \\ \frac{d y}{d x} = \frac{- (+ 3 m^{2} + 1)}{3 y^{2} + 2} \end{aligned}$
$\begin{aligned} m_{1} = \frac{- 1}{2} \\ x y + 2 x = y \\ \frac{d y}{d x} = \frac{- y - 2}{x - 1} \\ m_{2} = 2 \\ m_{1} m_{2} = - 1 \\ θ = 90^{0} \end{aligned}$

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An angle between the curvesx3+y3+x+2y=0,xy+2x=y at the origin is