Question

Show that the curves $x^3-3x{y}^2+2=0$ and $3x^{2}y-y-y^3-2=0$ cut orthogonally.

Answer 1

Given equation of the curves $x^3-3x{y}^2+2=0\cdots \left(1\right)$ and $3x^{2}y-y^3-2=0\cdots \left(2\right)$

Differentiating $\left(1\right)$ on both sides wrt $x$

$3x^2-3(y^2+2xy\frac{dy}{dx})+0=0$

$⟹{\left(\frac{dy}{dx}\right)}_1=\frac{3x^2-3y^2}{6xy}=\frac{x^2-y^2}{2xy}$

Differentiating $\left(2\right)$ on both sides wrt $x$

$3(2xy+x^2\frac{dy}{dx})-3y^2\frac{dy}{dx}-0=0$

$⟹{\left(\frac{dy}{dx}\right)}_2=\frac{-6xy}{3x^2-3y^2}=\frac{-2xy}{x^2-y^2}$

$⟹(\frac{dy}{dx}{)}_1(\frac{dy}{dx}{)}_2=-1$

$\frac{x^2-y^2}{2xy}\times$ $\frac{-2xy}{x^2-y^2}$ = $-1$

As the product of slopes of two tangents is $-1,$ the tangents are perpendicular to each other

Hence both the curves cut orthogonally