已知抛物线y=1/4x²+1和点A(0,2),点B是抛物线上一点,BC⊥x轴于点C,求证:AB=BC
设B点坐标为(x, y)
那么
y=1/4x^2+1
BC=y=1/4x^2+1
AB=√[(x-0)^2+(y-2)^2]
=√[x^2+(1/4x^2+1-2)^2]
=√[x^2+(1/4x^2-1)^2]
=√(x^2+1/16x^4-1/2x^2+1)
=√(1/16x^4+1/2x^2+1)
=√(1/4x^2+1)^2
=1/4x^2+1
=BC
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