Question

The two curves $x^3-3x{y}^2+2=0and3x^{2}y-y^3-2=0$

• A. Cut at right angles
• B. Touch each other
• C. Cut at an angle π/3
• D. Cut at an angle π/4

Answer 1

The correct option is A

Cut at right angles

$x^3-3x{y}^2+2=\mathrm{0...}\left(1\right)3x^{2}y-y^3-2=\mathrm{0...}\left(2\right)$
On differentiating equations (1) and (2) w.r.t x, we obtain
${\left(\frac{dy}{dx}\right)}_{c1}=\frac{x^2-y^2}{2xy}and{\left(\frac{dy}{dx}\right)}_{c2}=\frac{-2xy}{x^2-y^2}$
Since $m_1.m_2=-1$.Therefore the two curves cut at right angles.
Hence (a) is the correct answer.