CCTV100100
幼苗
共回答了17个问题采纳率:88.2% 举报
A(2,0),B(2,2),C(0,2)
抛物线顶点为(2,0),抛物线可表达为 y = a(x - 2)²
抛物线经过点C(0,2):2 = a(0 - 2)² = 4a,a = 1/2
y = (x - 2)²/2
(1) P(m,0),Q(0,6 -m)
OP = QM = m
OQ = PM = 6 - m
l = 2m+ 2(6-m) = 12
为常数
(2)
(i)m = 2时,P(2,0)在OA上或与A重合,Q(0,4)在BC上方.
0≤ < m ≤ 2时,重叠部分为正方形中的x 2
(4)二者均为直角三角形,只需一个锐角相等即可.
m = 3 - √5或m > 2
1年前
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CCTV100100
(3) 显然y = (x - 2)²/2过点D(4, 2) (i) 0 2) ∠APB = ∠BQC时, 2/(2-m) = 2/(4 - m), 无解 (2) 2 < m < 4 tan∠APB = AB/PA = 2/(m -2) tan∠QBC = CQ/CB = (6-m -2)/2 = (4 - m)/2 tan∠BQC = CB/CQ = 2/(6-m-2) = 2/(4 - m) ∠APB = ∠QBC时, 2/(m -2) = (4 - m)/2, 无解 ∠APB = ∠BQC时, 2/(m-2) = 2/(4-m), m = 3 (3) m > 4 tan∠APB = AB/PA = 2/(m -2) tan∠QBC = CQ/CB = (2-6+m)/2 = (m- 4)/2 tan∠BQC = CB/CQ = 2/(2 -6 +m) = 2/(m + 4) ∠APB = ∠QBC时, 2/(m -2) = (m -4)/2, m = 3 + √5 (舍去3 - √5 < 4) ∠APB = ∠BQC时, 2/(m-2) = 2/(m - 4), 无解 共有三个解: m = 3, m = 3 ± √5