Question

已知：矩形纸片ABCD中，AB=26厘米，BC=18.5厘米，点E在AD上，且AE=6厘米，点P是AB边上一动点．按如下操作：

步骤一，折叠纸片，使点P与点E重合，展开纸片得折痕MN（如图1所示）；

步骤二，过点P作PT⊥AB，交MN所在的直线于点Q，连接QE（如图2所示）

1.无论点P在AB边上任何位置，都有PQ_________QE（填“”、“”、“”号）；

2.如图3所示，将纸片ABCD放在直角坐标系中，按上述步骤一、二进行操作：

①当点P在A点时，PT与MN交于点Q₁，Q₁点的坐标是（_______，_________）；

②当PA=6厘米时，PT与MN交于点Q₂. Q₂点的坐标是（_______，_________）；

③当PA=12厘米时，在图3中画出MN,PT（不要求写画法），并求出MN与PT的交点Q₃的坐标；

3.点P在运动过程，PT与MN形成一系列的交点Q₁，Q₂，Q₃……观察、猜想：众多的交点形成的图象是什么？并直接写出该图象的函数表达式．

 

Answer 1

【答案】

 

1.

2.；②．③画图见解析。

3.抛物线 、函数关系式：

【解析】（1）．······························································································· 1分

（2）①；②．······························································································· 3分

③画图，如图所示．······································································································ 5分

解：方法一：设与交于点．

在中，，

．

，，

 

．

又，

．

．

．

．················································································································ 7分

方法二：过点作，垂足为，则四边形是矩形．

，．

设，则．

在中，．

．

．

．

．

（3）这些点形成的图象是一段抛物线．········································································ 8分

函数关系式：．····································································· 10分

说明：若考生的解答：图象是抛物线，函数关系式：均不扣分．