Q72 / 120 · VITEEE 2018 · Mathematics
The two curves $x^{3}-3 x y^{2}+2=0$ and $3 x^{2} y-y^{3}-2=0$ intersect at an angle of
- A$\frac{\pi}{4}$
- B$\frac{\pi}{3}$
- Correct answerC$\frac{\pi}{2}$
- D$\frac{\pi}{6}$
Solution
$x^{3}-3 x y^{2}+2=0$ differentiating w.r.t. $x$ :
$$
\begin{array}{l}
3 x^{2}-3 x(2 y) \frac{d y}{d x}-3 y^{2}=0 \\
\Rightarrow \frac{d y}{d x}=\frac{3 x^{2}-3 y^{2}}{6 x y} \text { and } 3 x^{2} y-y^{3}-2=0
\end{array}
$$
differentiating w.r.t. $\mathrm{x}$
$$
\begin{array}{l}
\Rightarrow 3 x^{2} \frac{d y}{d x}+6 x y-3 y^{2} \frac{d y}{d x}=0 \\
\Rightarrow \frac{d y}{d x}=-\left(\frac{6 x y}{3 x^{2}-3 y^{2}}\right)
\end{array}
$$
Now, product of slope
$$
=\frac{3 x^{2}-3 y^{2}}{6 x y} \times-\left(\frac{6 x y}{3 x^{2}-3 y^{2}}\right)=-1
$$
$\therefore$ they are perpendicular.
Hence, angle
$=\pi / 2$
Hence, the correct answer is option: 'C'.