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
1个回答
(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)
相关问题
