Question

实数x.y满足x平方+y平方-2x+4y=0。则x-2y的最大值

Answer 1

因为x^2+y^2-2x+4y=0
所以(x-1)^2+(y+2)^2=5
用参数方程得方法做
令x=√5cosθ+1,y=√5sinθ-2
则x-2y=√5cosθ+1-2(√5sinθ-2)
=5+√5cosθ-2√5sinθ
=5-5sin(θ-arctan(1/2))
所以x-2y的最大值是5-5*(-1)=10

Answer 2

x²+y²-2x+4y=0
∴(x-1)²+(y+2)²=5
用圆的参数方程
x=1+√5cost,y=-2+√5sint
x-2y=(1+√5cost)-2(-2+√5sint)=5+√5cost-2√5sint=5+5((√5/5)cost-(2√5/5)sint)
设sina=√5/5,cosa=2√5/5
∴x-2y=5+5(sinacost-sintcosa)=5+5sin(a-t)
∵-1<=sin(a-t)<=1
∴0<=5+5sin(a-t)<=10,即0<=x-2y<=10
最大值为10