若2X^2-6X+Y^2=0,求X^2-2AX+Y^2的最小值解:由2x^2-6x+y^2=0,得y^2=-2x^2+6x ≥0得出0≤x≤3,所以问题转化为求x^2-2ax+y^2=-x^2+2(3-a)x (0≤x≤3)的最小值;1.当a-3≤3/2,即a≤9/2时,(x^2-2ax+y^2)min=(-x^2+2(3-a)x)min=9-6
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