已知如图,Rt△ABC的两直角边OA,OB分别在x轴的正半轴和y轴的负半轴上,C为OA上一点

(1) x = p时,y = (p - 2)(p - m) - (p - 2)(p - m) = 0,所以(p,0)在抛物线上.
(2) 设A(a,0),C(c,0)
x = a时,y = (a - 2)(a - m) - (p - 2)(p - m) = 0 (1)
x = c时,y = (c - 2)(c - m) - (p - 2)(p - m) = 0 (2)
(1)-(2):(a - 2)(a - m) - (c - 2)(c - m) = 0
a² - (m+2)a + 2m = c² - (m+2)c + 2m
a² - c² = (m+2)a - (m+2)c
(a + c)(a - c) = (m + 2)(a - c)
根据题意,a,c不相等,a + c = m + 2
因为抛物线与x轴只能有两个交点,(p,0)必为A或C.
(a) (p,0)为A
a = p
c = m + 2 - a = m + 2 - p
根据题意,a > c,m + 2 ≥ 2p
a -c = p - (m + 2 - p) = 2p - (m + 2) < 0
与a > c矛盾,舍去
(b) (p,0)为C
c = p
a = m + 2 - c = m + 2 - p
根据题意,a > c,m + 2 ≥ 2p
a - c = (m + 2 - p) - p = (m + 2) - 2p > 0
符合题意
|OC| = c = p
|OA| = a = m + 2 - p
(3)|OB| = |OC| = p
△AOB的面积S = (1/2)|OB|*|OA| = p(m+2-p)/2
= -[p - (m+2)/2]² + (m+2)²/8
p = (m+2)/2时,S最大非常感谢你的帮助！请问你有没有一些优录的初中试题呢？语数外理化（有几科给几科）最好都有、有吗？如果有的话请发到我的邮箱625284078@qq.com、我会提高悬赏的！谢谢！抱歉，我没有这类材料。