已知在平面直角坐标系中,A、B两点在x轴上,线段OA、OB的长分别为方程x²-8x+12=0的两个根,点C是y

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  • (1) x² = 8x + 12 = 0

    (x-2)(x-6) = 0

    x = 2或x = 6

    A(2,0),B(6,0)

    (2)过A、B的抛物线可表示为y = a(x - 2)(x - 6)

    取x = 0,-3 = 12a

    a = -1/4

    y = -(x - 2)(x - 6)/4

    (3)对称轴:x = (6+2)/2 = 4

    此时y = -(4 - 2)(4 - 6)/4 = 1

    E(4,1)

    题中有遗漏,D来历不明,估计是C关于对称轴的对称点,以下按此做.

    CE = √[(4-0)² + (1+3)²] = 4√2

    ①当△CEM是等膘三角形时

    (i) CE = CM且M在C上方

    M(0,4√2 - 3)

    (ii) CE = CM且M在C下方

    M(0,-4√2 - 3)

    (iii) CM = EM

    设M(0,m)

    CM = |m +3|

    EM = √[(4-0)² + (m - 1)²] = √[16 + (m - 1)²]

    √[16 + (m - 1)²] = |m +3|

    两边平方,解得m = 1

    M(0,1)

    (iv) CE = EM

    设M(0,m)

    CE = 4√2

    EM = √[(4-0)² + (m - 1)²] = √[16 + (m - 1)²]

    √[16 + (m - 1)²] = 4√2

    两边平方,解得m = 5或m = -3(此为C,舍去)

    M(0,5)