Question

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

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

Answer 1

The correct option is B Cut each other at right angle

Consider the equation $x^3-3x{y}^2+2=0$ and differentiate it with respect to $x$:

$3x^2-3y^2-6xy\frac{dy}{dx}=0\Rightarrow 6xy\frac{dy}{dx}=3x^2-3y^2\Rightarrow \frac{dy}{dx}=\frac{3x^2-3y^2}{6xy}\Rightarrow \frac{dy}{dx}=\frac{x^2-y^2}{2xy}.....\left(1\right)$

Now, consider the equation $3x^{2}y-y^3-2=0$ and differentiate it with respect to $x$:

$6xy+3x^2\frac{dy}{dx}-3y^2\frac{dy}{dx}=0\Rightarrow 6xy+\frac{dy}{dx}(3x^2-3y^2)=0\Rightarrow \frac{dy}{dx}=\frac{-6xy}{3x^2-3y^2}\Rightarrow \frac{dy}{dx}=\frac{-2xy}{x^2-y^2}.....\left(2\right)$

Multiply 1 and 2 as follows:

$\left(\frac{x^2-y^2}{2xy}\right)\left(\frac{-2xy}{x^2-y^2}\right)=-1$

In general, if the product of two slopes is equal to $-1$ then the lines are perpendicular, therefore, the two given curves are perpendicular to each other.

Hence, the two curves $x^3-3x{y}^2+2=0$ and $3x^{2}y-y^3-2=0$ cut each other at right angle.