Question

The two curves $x^3-3x{y}^2+2=0$ and $3x^{2}y-y^3-2=0$ intersect at an angle of

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (C)

$x^3-3x{y}^2+2=0$
differentiating w.r.t. x
$3x^2-3x\left(2y\right)\frac{dy}{dx}-3y^2=0$
$\Rightarrow \frac{dy}{dx}=\frac{3x^2-3y^2}{6xy}$ and $3x^{2}y-y^3-2=0$
differentiating w.r.t. x
$3x^2\frac{dy}{dx}+6xy-3y^2\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=-\left(\frac{6xy}{3x^2-3y^2}\right)$
Now, product of slope
$=\frac{3x^2-3y^2}{6xy}\times -\left(\frac{6xy}{3x^2-3y^2}\right)=1$
$\therefore$ they are perpendicular.
Hence, angle
$=\pi /2$