设实数x,y满足(x-1)^2+(y+2)^2=5,则x-2y的最大值
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设实数x,y满足(x-1)^2+(y+2)^2=5,则x-2y的最大值设实数x,y满足(x-1)^2+(y+2)^2=5,则x-2y的最大值设实数x,y满足(x-1)^2+(y+2)^2=5,则x-2
设实数x,y满足(x-1)^2+(y+2)^2=5,则x-2y的最大值
设实数x,y满足(x-1)^2+(y+2)^2=5,则x-2y的最大值
设实数x,y满足(x-1)^2+(y+2)^2=5,则x-2y的最大值
实数x,y满足(x-1)^2+(y+2)^2=5
设为参数方程为
x=1+√5cosθ
y=-2+√5sinθ
x-2y =1+√5cosθ-2(-2+√5sinθ)
=5-(2√5sinθ-√5cosθ)
=5-√(2√5^2+√5^2)
=5-5sin(θ+φ)
要值最大,则sin(θ+φ)最小为-1
(x-2y)max=10
(x-1)^2+(y+2)^2=5
参数方程:
x=√5cost+1,y=√5sint-2
假设:cota=2,1/sina=√5/2
x-2y
=√5cost-2√5sint+5
=√5(cost-2sint)+5
=√5(cost-cota*sint)+5
=√5*1/sina*(sina*cost-cosa*sint)+5
=5/2*sin(a-t)+5
sin(a-t)=1时
(x-2y)max=15/2
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