Question

Find the min value of f(x, y) = x² + y², where x²(xy – 1) = y²(1 + xy)

Answer 1

Put x = r cosθ, y = rsinθ

We have x²(xy – 1) = y²(1 + xy)

⇒  xy(x² – y²) = x² + y²

⇒ r⁴sinθ. cosθ. cos 2θ = r²

⇒  r²(2sinθ. cosθ). cos2θ = 2

⇒  r²(2sin2θcos2θ) = 4

⇒  r²(sin4θ) = 4

⇒ r² = 4/sin4θ = 4cosec4θ

⇒f(x, y) = x² + y²

Thus, f(x, y) = x² + y² 

 = r²

= 4cosec4θ

Min value is 4.