Prove that the curves xy = 4 and x² + y² = 8 touch each other.

Given:

Curves xy = 4 ...(1)

& x² + y² = 8 ...(2)

Solving (1) & (2),we get

⇒ xy = 4

⇒ x

Substituting x in x² + y² = 8,we get,

⇒ ()² + y² = 8

⇒ + y² = 8

⇒ 16 + y⁴ = 8y²

⇒ y⁴ – 8y² + 16 = 0

We will use factorization method to solve the above equation

⇒ y⁴ – 4y² – 4y² + 16 = 0

⇒ y²(y² – 4) – 4(y² – 4) = 0

⇒ (y² – 4)(y² – 4) = 0

⇒ y² – 4 = 0

⇒ y² = 4

⇒ y = ±2

Substituting y = ±2 in x,we get,

⇒ x

⇒ x = ±2

∴ The point of intersection of two curves (2,2) &

( – 2, – 2)

First curve xy = 4

⇒ 1×y + x. = 0

⇒ x. = – y

⇒ m₁ ...(3)

Second curve is x² + y² = 8

Differentiating above w.r.t x,

⇒ 2x + 2y. = 0

⇒ y. = – x

⇒ m₂ ...(4)

At (2,2),we have,

m₁

⇒

m₁ = – 1

At (2,2),we have,

⇒ m₂

⇒

⇒ m₂ = – 1

Clearly, m_{1 =} m₂ = – 1 at (2,2)

So, given curve touch each other at (2,2)