26．解:(1)由已知.得.. . ． ．············································································································ 設(shè)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(shè)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(shè)． ..． 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(nèi)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(shè). ． ． 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(jīng)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(jié)． (1)求該拋物線的解析式, (2)若動點從點出發(fā).以每秒1個長度單位的速度沿射線運動.設(shè)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(fā).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(shè)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(jīng)過點. ·························································································· 1分 二次函數(shù)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】

 

題目列表(包括答案和解析)

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(shè)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(guān)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(jié)論是否成立？并說明理由；
（3）在（2）的條件下，設(shè)
BP
PD
=k，是否存在這樣的實數(shù)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．
精英家教網(wǎng)

查看答案和解析>>

已知：在平面直角坐標系xOy中，二次函數(shù)y=x2+bx+c的圖象與x軸交于A、B兩點，點A在點B的左側(cè)，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(shù)的解析式；
（2）設(shè)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．

查看答案和解析>>

已知：在平面直角坐標系xOy中，二次函數(shù)y=x2+bx+c的圖象與x軸交于A、B兩點，點A在點B的左側(cè)，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(shù)的解析式；
（2）設(shè)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．

查看答案和解析>>

已知：在平面直角坐標系xOy中，二次函數(shù)y=x2+bx+c的圖象與x軸交于A、B兩點，點A在點B的左側(cè)，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(shù)的解析式；
（2）設(shè)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．

查看答案和解析>>

已知拋物線y＝x2＋4x＋m(m為常數(shù))經(jīng)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(shè)為直線l2)與平移前的拋物線的對稱軸(設(shè)為直線l1)關(guān)于y軸對稱；它所對應的函數(shù)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．

查看答案和解析>>


同步練習冊答案