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Q.
The two curves $x^3-3xy^2+2=0$ and $3x^2y-y^3-2=0$
Application of Derivatives
Solution:
$x^{3}-3xy^{2}+2=0$
$\Rightarrow 3x^{2}-3y^{2}-6\,xy \frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{3\left(x^{2}-y^{2}\right)}{6\,xy}=\frac{x^{2}-y^{2}}{2\,xy}$
Again $3x^{2}y-y^{2}-2=0$
$\Rightarrow 3x^{2} \frac{dy}{dx}+6\,xy-3y^{2} \frac{dy}{dx}=0$
$\Rightarrow 3\left(x^{2}-y^{2}\right) \frac{dy}{dx}=-6xy$
$\Rightarrow \frac{dy}{dx}=-\frac{2xy}{x^{2}-y^{2}}$ Let $\left(x_{1}, y_{1}\right)$ be the point of intersection of two curves.
$\therefore $ product of slopes of tangents at $\left(x_{1}, y_{1}\right)$
$=\frac{x^{2}_{1}-y^{2}_{1}}{2\,x_{1}y_{1}}\cdot\frac{-2\,x_{1}y_{1}}{x^{2}_{1}-y^{2}_{1}}=-1$
$\therefore $ curves cut at right angles.