# Two curves $x^3-3xy^2+2=0$ and $3x^2-y^3-2=0$ intersect at an angle of

$(A)\;\large\frac{\pi}{4}\quad(B)\;\large\frac{\pi}{3}\quad(C)\;\large\frac{\pi}{2}\quad(D)\;\large\frac{\pi}{6}$

Toolbox:
• Slope of a line is $ax+by+c=0$ is $- \bigg( \large\frac{ Coefficient\: of \: x}{ Coefficient\: of \: y} \bigg)$
• If $y = f(x),\: then \: \bigg( \large\frac{dy}{dx} \bigg)_p =$ slope of the normal to $y = f(x)$ at point $p$
Step 1
The given curves are $x^3-3xy^2+2=0\: and \: 3x^2y-y^3-2=0$
Consider the equation
$x^3-3xy^2+2=0$
differentiating w.r.t $x$ we get,
$3x^2-(3x.2y.\large\frac{dy}{dx}+y^2.3)+0=0$
$\Rightarrow 3x^2-6xy.\large\frac{dy}{dx}-3y^2=0$
$\therefore \large\frac{dy}{dx}=\large\frac{3y^2-3x^2}{-6xy}$
$-\large\frac{(y^2-x^2)}{2xy}$
Let this be $m_1$
Step 2
Consider the equation
$3x^2-y^3-2=0$
differentiating w.r.t $x$ we get,
$3x^2.\large\frac{dy}{dx}+y.6x-3y^2.\large\frac{dy}{dx}=0$
$\Rightarrow x^2.\large\frac{dy}{dx}+2xy-y^2.\large\frac{dy}{dx}=0$
$\Rightarrow \large\frac{dy}{dx}(x^2-y^2)=-2xy$
$\Rightarrow \large\frac{dy}{dx}=\large\frac{-2xy}{x^2-y^2}$
Let this be $m_2$
The product of the slopes is $m_1 \times m_2$
$= -\bigg( \large\frac{y^2-x^2}{2xy} \bigg) \times \large\frac{-2xy}{x^2-y^2}$
$= -1$
Since the product of the slopes is $-1$, the angle between them is $\large\frac{\pi}{2}$
Hence C is the correct option.