Show that the following set of curves intersect orthogonally :

x³ – 3xy² = – 2 and 3x² y – y³ = 2

Given:

Curves x³ – 3xy² = – 2 ...(1)

& 3x²y – y³ = 2 ...(2)

Adding (1) & (2),we get

⇒ x³ – 3xy² + 3x²y – y³ = – 2 + 2

⇒ x³ – 3xy² + 3x²y – y³ = – 0

⇒ (x – y)³ = 0

⇒ (x – y) = 0

⇒ x = y

Substituting x = y on x³ – 3xy² = – 2

⇒ x³ – 3×x×x² = – 2

⇒ x³ – 3x³ = – 2

⇒ – 2x³ = – 2

⇒ x³ = 1

⇒ x = 1

Since x = y

y = 1

The point of intersection of two curves is (1,1)

First curve x³ – 3xy² = – 2

Differentiating above w.r.t x,

⇒ 3x² – 3(1×y² + x×2y) = 0

⇒ 3x² – 3y² – 6xy0

⇒ 3x² – 3y² = 6xy

⇒

⇒

⇒ m₁ ...(3)

Second curve 3x²y – y³ = 2

Differentiating above w.r.t x,

⇒ 3(2x×y + x²×) – 3y²0

⇒ 6xy + 3x²3y²0

⇒ 6xy + (3x² – 3y²)0

⇒

⇒

⇒ m₂ ...(4)

When m₁ & m₂ =

⇒ × = – 1

∴ Two curves x³ – 3xy² = – 2 & 3x²y – y³ = 2 intersect orthogonally.