26．解:(1)由已知.得.. . ．．············································································································ 設過點的拋物線的解析式為．將點的坐標代入.得．將和點的坐標分別代入.得 ··································································································· 解這個方程組.得故拋物線的解析式為．··························································· (2)成立．························································································· 點在該拋物線上.且它的橫坐標為. 點的縱坐標為．······················································································· 設的解析式為. 將點的坐標分別代入.得解得的解析式為．········································································ .．··························································································· 過點作于點. 則． . ．又. ．．．··········································································································· ． (3)點在上...則設． ..． ①若.則. 解得．.此時點與點重合．．··········································································································· ②若.則. 解得 ..此時軸．與該拋物線在第一象限內的交點的橫坐標為1. 點的縱坐標為．．······································································································· ③若.則. 解得..此時.是等腰直角三角形．過點作軸于點. 則.設. ．．解得．．··········································· 綜上所述.存在三個滿足條件的點. 即或或． 26．如圖.已知拋物線經過點.拋物線的頂點為.過作射線．過頂點平行于軸的直線交射線于點.在軸正半軸上.連結． (1)求該拋物線的解析式, (2)若動點從點出發.以每秒1個長度單位的速度沿射線運動.設點運動的時間為．問當為何值時.四邊形分別為平行四邊形?直角梯形?等腰梯形? (3)若.動點和動點分別從點和點同時出發.分別以每秒1個長度單位和2個長度單位的速度沿和運動.當其中一個點停止運動時另一個點也隨之停止運動．設它們的運動的時間為.連接.當為何值時.四邊形的面積最小?并求出最小值及此時的長． *26．解:(1)拋物線經過點. ·························································································· 1分二次函數的解析式為:·················································· 3分 (2)為拋物線的頂點過作于.則. ··················································· 4分當時.四邊形是平行四邊形 ················································ 5分當時.四邊形是直角梯形過作于.則 (如果沒求出可由求) ····························································································· 6分當時.四邊形是等腰梯形綜上所述:當.5.4時.對應四邊形分別是平行四邊形.直角梯形.等腰梯形．·· 7分及已知.是等邊三角形則過作于.則········································································· 8分 =·································································································· 9分當時.的面積最小值為··································································· 10分此時 ······················································ 11分【查看更多】

已知四邊形ABCD中，P是對角線BD上的一點，過P作MN∥AD，EF∥CD，分別交AB、CD、AD、BC于點M、N、E、F，設a=PM•PE，b=PN•PF，解答下列問題：
（1）當四邊形ABCD是矩形時，見圖1，請判斷a與b的大小關系，并說明理由；
（2）當四邊形ABCD是平行四邊形，且∠A為銳角時，見圖2，（1）中的結論是否成立？并說明理由；
（3）在（2）的條件下，設

BP

PD

=k，是否存在這樣的實數k，使得

S平行四邊形PEAM

S△ABD

=

4

9

？若存在，請求出滿足條件的所有k的值；若不存在，請說明理由．

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已知：在平面直角坐標系xOy中，二次函數y=x²+bx+c的圖象與x軸交于A、B兩點，點A在點B的左側，若拋物線的對稱軸為x=1，點A的坐標為（-1，0）．
（1）求這個二次函數的解析式；
（2）設拋物線的頂點為C，拋物線上一點D的坐標為（-3，12），過點B、D的直線與拋物線的對稱軸交于點E．問：是否存在這樣的點F，使得以點B、C、E、F為頂點的四邊形是平行四邊形？若存在，求出點F的坐標；若不存在，請說明理由；
（3）在（2）的條件下，若在BD上存在一點P，使得直線AP將四邊形ACBD分成了面積相等的兩部分，請你求出此時點P的坐標．

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已知：在平面直角坐標系xOy中，二次函數y=x²+bx+c的圖象與x軸交于A、B兩點，點A在點B的左側，若拋物線的對稱軸為x=1，點A的坐標為（-1，0）．
（1）求這個二次函數的解析式；
（2）設拋物線的頂點為C，拋物線上一點D的坐標為（-3，12），過點B、D的直線與拋物線的對稱軸交于點E．問：是否存在這樣的點F，使得以點B、C、E、F為頂點的四邊形是平行四邊形？若存在，求出點F的坐標；若不存在，請說明理由；
（3）在（2）的條件下，若在BD上存在一點P，使得直線AP將四邊形ACBD分成了面積相等的兩部分，請你求出此時點P的坐標．

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已知拋物線y＝x²＋4x＋m(m為常數)經過點(0，4)．

(1)求m的值；

(2)將該拋物線先向右、再向下平移得到另一條拋物線．已知平移后的拋物線滿足下述兩個條件：它的對稱軸(設為直線l₂)與平移前的拋物線的對稱軸(設為直線l₁)關于y軸對稱；它所對應的函數的最小值為－8．

①試求平移后的拋物線的解析式；

②試問在平移后的拋物線上是否存在點P，使得以3為半徑的圓P既與x軸相切，又與直線l₂相交？若存在，請求出點P的坐標，并求出直線l₂被圓P所截得的弦AB的長度；若不存在，請說明理由．

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