Question

Show that the following set of curve intersect orthogonally x³ − 3xy² = −2 and 3x²y − y³ = 2 ?

Answer 1

\[\text { Let the given curves intersect at }\left( x_1 , y_1 \right)\]

\[ x^3 - 3x y^2 = - 2 . . . \left( 1 \right)\]

\[3 x^2 y - y^3 = 2 . . . \left( 2 \right)\]

\[\text { Differentiating (1) w.r.t.x,}\]

\[3 x^2 - 3 y^2 - 6xy\frac{dy}{dx} = 0\]

\[ \Rightarrow \frac{dy}{dx} = \frac{3 x^2 - 3 y^2}{6xy} = \frac{x^2 - y^2}{2xy}\]

\[ \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) = \frac{{x_1}^2 - {y_1}^2}{2 x_1 y_1}\]

\[\text { Differenntiating (2) w.r.t.x, }\]

\[3 x^2 \frac{dy}{dx} + 6xy - 3 y^2 \frac{dy}{dx} = 0\]

\[ \Rightarrow \frac{dy}{dx}\left( 3 x^2 - 3 y^2 \right) = - 6xy\]

\[ \Rightarrow \frac{dy}{dx} = \frac{- 6xy}{3 x^2 - 3 y^2} = \frac{- 2xy}{x^2 - y^2}\]

\[ \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) = \frac{- 2 x_1 y_1}{{x_1}^2 - {y_1}^2}\]

\[\text { Now,} m_1 \times m_2 = \frac{{x_1}^2 - {y_1}^2}{2 x_1 y_1} \times \frac{- 2 x_1 y_1}{{x_1}^2 - {y_1}^2}\]

\[ \Rightarrow m_1 \times m_2 = - 1\]

\[Since, m_1 \times m_2 = - 1\]

So, the given curves intersect orthogonally.