摘要：26．(1)解:由得点坐标为 由得点坐标为 ∴··················································································· 由解得∴点的坐标为···································· ∴··························································· (2)解:∵点在上且 ∴点坐标为······················································································ 又∵点在上且 ∴点坐标为······················································································ ∴··········································································· (3)解法一:当时.如图1.矩形与重叠部分为五边形(时.为四边形)．过作于.则 ∴即∴ ∴ 即···································································
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解：（1）OA=1，OC=2
则A点坐标为（0，1），C点坐标为（2，0）
设直线AC的解析式为y=kx+b
解得
直线AC的解析式为
··················· 2分
（2）
或
（正确一个得2分）························· 8分
（3）如图，设
过
点作
于F
由折叠知
或2··········· 10分
在直角坐标系xOy中，设点A（0，t），点Q（t，b）（t，b均为非零常数）．平移二次
函数y=-tx2的图象，得到的抛物线F满足两个条件：①顶点为Q；②与x轴相交于B，C两点（|OB|＜|OC|）．连接AB．
（1）是否存在这样的抛物线F，使得|OA|2=|OB|•|OC|？请你作出判断，并说明理由；
（2）如果AQ∥BC，且tan∠ABO=
，求抛物线F对应的二次函数的解析式．
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函数y=-tx2的图象，得到的抛物线F满足两个条件：①顶点为Q；②与x轴相交于B，C两点（|OB|＜|OC|）．连接AB．（1）是否存在这样的抛物线F，使得|OA|2=|OB|•|OC|？请你作出判断，并说明理由；
（2）如果AQ∥BC，且tan∠ABO=
| 3 | 2 |
函数y=-tx2的图象，得到的抛物线F满足两个条件：①顶点为Q；②与x轴相交于B，C两点（|OB|＜|OC|）．连接AB．
，求抛物线F对应的二次函数的解析式．
，求抛物线F对应的二次函数的解析式。
的图象，得到的抛物线F满足两个条件：①顶点为Q；②与x轴相交于B，C两点（OB<OC），连结A，B。
？请你做出判断，并说明理由；
，求抛物线F对应的二次函数的解析式。