Question

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

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

Answer 1

Answer: (A)

$x^3-3x{y}^2+2=0$
$\Rightarrow 3x^2-3y^2-6xy\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{3\left(x^2-y^2\right)}{6xy}=\frac{x^2-y^2}{2xy}$
Again $3x^{2}y-y^2-2=0$
$\Rightarrow 3x^2\frac{dy}{dx}+6xy-3y^2\frac{dy}{dx}=0$
$\Rightarrow 3\left(x^2-y^2\right)\frac{dy}{dx}=-6xy$
$\Rightarrow \frac{dy}{dx}=-\frac{2xy}{x^2-y^2}$
Let $\left(x_1,y_1\right)$ be the point of intersection of two curves.
$\therefore$ product of slopes of tangents at $\left(x_1,y_1\right)$
$=\frac{x_1^2-y_1^2}{2x_1{y}_1}\cdot \frac{-2x_1{y}_1}{x_1^2-y_1^2}=-1$
$\therefore$ curves cut at right angles.