近年高中数学联赛有这样一道题 :实数x ,y满足 4x2 - 5xy +4 y2 =5,设S =x2 +y2 ,则 1Smax+1Smin的值为 .下面给出这道题的多种解法 .解法 1　由题设易知S =x2 +y2 0 ,设x =Scosθy =Ssinθθ为参数 ,代入 4x2 - 5xy+4y2 =5,得 4Scos2 θ- 5Ssinθcosθ +4Ssinθ=5,所以sin2θ In recent years, the high school mathematics league has such a question: the real number x, y satisfies 4x2 - 5xy +4 y2 =5, and if S =x2 +y2, then 1Smax + 1Smin is the value. Various solutions to this problem are given below. 1 Let the problem be known as S = x2 + y2 0 , let x = Scosθy = Ssinθθ be a parameter and substitute 4x2 - 5xy + 4y2 = 5 to get 4Scos2 θ - 5Ssinθcosθ + 4Ssinθ = 5, so sin2θ