实数x,y满足3x^2+2y^2=6x,求x^2+y^2的最小值和最大值高中参数方程过程
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实数x,y满足3x^2+2y^2=6x,求x^2+y^2的最小值和最大值高中参数方程过程
实数x,y满足3x^2+2y^2=6x,求x^2+y^2的最小值和最大值高中参数方程过程
实数x,y满足3x^2+2y^2=6x,求x^2+y^2的最小值和最大值高中参数方程过程
3x^2+2y^2=6x
3x^2-6x+3+2y^2=3
3(x-1)^2+2y^2=3
(x-1)^2+(2/3)y^2=1
令x-1=sina y=√(3/2)cosa
x^2+y^2
=(1+sina)^2+(3/2)(cosa)^2
=(sina)^2+2sina+1+(cosa)^2+(1/2)[1-(sina)^2]
=-(sina)^2/2+2sina+5/2
=(-1/2)[(sina)^2-4sina-5]
=(-1/2)(sina-2)^2+9/2
当sina=-1时,x^2+y^2有最小值0
当sina=1时,x^2+y^2有最大值4
3x^2+2y^2=6x
--->3(x^2-2x)^2+2y^2=0
--->3(x-1)^2+2y^2=3
--->(x-1)^2+y^2/(3/2)=1......(*)
--->x=1+cost; y=√(3/2)sint
因此x^2+y^2=(1+cost)^2+3/2*(sint)^2
=[1+2cost+(cost)^2]+...
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3x^2+2y^2=6x
--->3(x^2-2x)^2+2y^2=0
--->3(x-1)^2+2y^2=3
--->(x-1)^2+y^2/(3/2)=1......(*)
--->x=1+cost; y=√(3/2)sint
因此x^2+y^2=(1+cost)^2+3/2*(sint)^2
=[1+2cost+(cost)^2]+3/2[1-(cost)^2]
=-1/2*(cost)^2+2cost+5/2
=-1/2*(cost-2)^2+9/2
-1=
--->-9/2=<-1/2*(cost)^2=<-1/2
--->0=<-1/2*(cost-2)^2+9/2=<4
--->0=
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