Mathematics - Application of Derivative Question with Solution | TestHub

$x^3-3x{y}^2+2=0$ differentiating w.r.t. $x$ : $$\begin{array}{l}3x^2-3x(2y)\frac{dy}{dx}-3y^2=0\\ \Rightarrow \frac{dy}{dx}=\frac{3x^2-3y^2}{6xy}\text{ and }3x^{2}y-y^3-2=0\end{array}$$ differentiating w.r.t. $\mathrm{x}$ $$\begin{array}{l}\Rightarrow 3x^2\frac{dy}{dx}+6xy-3y^2\frac{dy}{dx}=0\\ \Rightarrow \frac{dy}{dx}=-\left(\frac{6xy}{3x^2-3y^2}\right)\end{array}$$ 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 $=\pi /2$