Question

The two curves $x^3 - 3xy^2 + 2 = 0$ and $3x^2y - y^3 - 2 = 0$ 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)

$x^{3} - 3xy^{2} + 2 = 0$
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
$= \pi/2$