Question

已知：矩形紙片ABCD中，AB=26厘米，BC=18.5厘米，點(diǎn)E在AD上，且AE=6厘米，點(diǎn)P是AB邊上一動(dòng)點(diǎn)．按如下操作：

步驟一，折疊紙片，使點(diǎn)P與點(diǎn)E重合，展開(kāi)紙片得折痕MN（如圖1所示）；

步驟二，過(guò)點(diǎn)P作PT⊥AB，交MN所在的直線于點(diǎn)Q，連接QE（如圖2所示）

1.無(wú)論點(diǎn)P在AB邊上任何位置，都有PQ_________QE（填“”、“”、“”號(hào)）；

2.如圖3所示，將紙片ABCD放在直角坐標(biāo)系中，按上述步驟一、二進(jìn)行操作：

①當(dāng)點(diǎn)P在A點(diǎn)時(shí)，PT與MN交于點(diǎn)Q₁，Q₁點(diǎn)的坐標(biāo)是（_______，_________）；

②當(dāng)PA=6厘米時(shí)，PT與MN交于點(diǎn)Q₂. Q₂點(diǎn)的坐標(biāo)是（_______，_________）；

③當(dāng)PA=12厘米時(shí)，在圖3中畫(huà)出MN,PT（不要求寫(xiě)畫(huà)法），并求出MN與PT的交點(diǎn)Q₃的坐標(biāo)；

3.點(diǎn)P在運(yùn)動(dòng)過(guò)程，PT與MN形成一系列的交點(diǎn)Q₁，Q₂，Q₃……觀察、猜想：眾多的交點(diǎn)形成的圖象是什么？并直接寫(xiě)出該圖象的函數(shù)表達(dá)式．

 

Answer 1

 

1.

2.；②．③畫(huà)圖見(jiàn)解析。

3.拋物線 、函數(shù)關(guān)系式：

解析:（1）．······························································································· 1分

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

③畫(huà)圖，如圖所示．······································································································ 5分

解：方法一：設(shè)與交于點(diǎn)．

在中，，

．

，，

 

．

又，

．

．

．

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

方法二：過(guò)點(diǎn)作，垂足為，則四邊形是矩形．

，．

設(shè)，則．

在中，．

．

．

．

．

（3）這些點(diǎn)形成的圖象是一段拋物線．········································································ 8分

函數(shù)關(guān)系式：．····································································· 10分

說(shuō)明：若考生的解答：圖象是拋物線，函數(shù)關(guān)系式：均不扣分．