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探索研究如圖.在直角坐標(biāo)系中.點為函數(shù)在第一象限內(nèi)的圖象上的任一點.點的坐標(biāo)為.直線過且與軸平行.過作軸的平行線分別交軸.于.連結(jié)交軸于.直線交軸于． (1)求證:點為線段的中點, (2)求證:①四邊形為平行四邊形, ②平行四邊形為菱形, (3)除點外.直線與拋物線有無其它公共點?并說明理由． (1)法一:由題可知． .. ．························································································· .即為的中點．····································································· 法二:..．·························································· 又軸.．··············································································· 可知.. .. ．·························································································· . 又.四邊形為平行四邊形．···················································· ②設(shè).軸.則.則．過作軸.垂足為.在中. ．平行四邊形為菱形．··············································································· (3)設(shè)直線為.由.得.代入得: 直線為．························ 設(shè)直線與拋物線的公共點為.代入直線關(guān)系式得: ..解得．得公共點為．所以直線與拋物線只有一個公共點．············································· 80 如圖.在平面直角坐標(biāo)系中.點.點分別在軸.軸的正半軸上.且滿足． (1)求點.點的坐標(biāo)． (2)若點從點出發(fā).以每秒1個單位的速度沿射線運動.連結(jié)．設(shè)的面積為.點的運動時間為秒.求與的函數(shù)關(guān)系式.并寫出自變量的取值范圍．的條件下.是否存在點.使以點為頂點的三角形與相似?若存在.請直接寫出點的坐標(biāo),若不存在.請說明理由． (08黑龍江齊齊哈爾28題解析)解:(1) .··················································································· . 點.點分別在軸.軸的正半軸上 ······························································································· (2)求得························································································· (每個解析式各1分.兩個取值范圍共1分)························································· (3),,, ···························································································································· 81如圖13.已知拋物線經(jīng)過原點O和x軸上另一點A,它的對稱軸x=2 與x軸交于點C.直線y=-2x-1經(jīng)過拋物線上一點B(-2,m).且與y軸.直線x=2分別交于點D.E. (1)求m的值及該拋物線對應(yīng)的函數(shù)關(guān)系式,(2)求證:① CB=CE ,② D是BE的中點, (3)若P(x.y)是該拋物線上的一個動點.是否存在這樣的點P,使得PB=PE,若存在.試求出所有符合條件的點P的坐標(biāo),若不存在.請說明理由. (1)∵ 點B(-2,m)在直線y=-2x-1上. ∴ m=-2×(-2)-1=3. ------------ ∴ B ∵ 拋物線經(jīng)過原點O和點A.對稱軸為x=2. ∴ 點A的坐標(biāo)為(4,0) . 設(shè)所求的拋物線對應(yīng)函數(shù)關(guān)系式為y=a(x-0)(x-4). -------- 將點B代入上式.得3=a.∴ . ∴ 所求的拋物線對應(yīng)的函數(shù)關(guān)系式為.即. (2)①直線y=-2x-1與y軸.直線x=2的交點坐標(biāo)分別為D E. 過點B作BG∥x軸.與y軸交于F.直線x=2交于G. 則BG⊥直線x=2.BG=4. 在Rt△BGC中.BC=. ∵ CE=5. ∴ CB=CE=5. -------- ②過點E作EH∥x軸.交y軸于H. 則點H的坐標(biāo)為H. 又點F.D的坐標(biāo)為F(0,3).D. ∴ FD=DH=4.BF=EH=2.∠BFD=∠EHD=90°. ∴ △DFB≌△DHE (SAS). ∴ BD=DE. 即D是BE的中點. ------------ (3) 存在. ------------ 由于PB=PE.∴ 點P在直線CD上. ∴ 符合條件的點P是直線CD與該拋物線的交點. 設(shè)直線CD對應(yīng)的函數(shù)關(guān)系式為y=kx+b. 將D C(2,0)代入.得. 解得 . ∴ 直線CD對應(yīng)的函數(shù)關(guān)系式為y=x-1. ∵ 動點P的坐標(biāo)為(x.). ∴ x-1=. ------------ 解得 .. ∴ .. ∴ 符合條件的點P的坐標(biāo)為(.)或(.).- 【查看更多】

題目列表(包括答案和解析)

（2010江蘇鎮(zhèn)江）探索發(fā)現(xiàn)（本小題滿分9分）

如圖，在直角坐標(biāo)系的直角頂點A，C始終在x軸的正半軸上，B，D在第一象限內(nèi)，點B在直線OD上方，OC=CD，OD=2，M為OD的中點，AB與OD相交于E，當(dāng)點B位置變化時，

試解決下列問題：

（1）填空：點D坐標(biāo)為；

（2）設(shè)點B橫坐標(biāo)為t，請把BD長表示成關(guān)于t的函數(shù)關(guān)系式，并化簡；

（3）等式BO=BD能否成立？為什么？

（4）設(shè)CM與AB相交于F，當(dāng)△BDE為直角三角形時，判斷四邊形BDCF的形狀，并證明你的結(jié)論.

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（2010 江蘇鎮(zhèn)江）推理證明（本小題滿分7分）

如圖，已知△ABC中，AB=BC，以AB為直徑的⊙O交AC于點D，過D作DE⊥BC，垂足為E，連結(jié)OE，CD=，∠ACB=30°.

（1）求證：DE是⊙O的切線；

（2）分別求AB，OE的長；

（3）填空：如果以點E為圓心，r為半徑的圓上總存在不同的兩點到點O的距離為1，則r的取值范圍為 .

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如圖，已知一次函數(shù)y = - x +7與正比例函數(shù)y = x的圖象交于點A，且與x軸交于點B.

（1）求點A和點B的坐標(biāo)；

（2）過點A作AC⊥y軸于點C，過點B作直線l∥y軸．動點P從原點O出發(fā)，以每秒1個單位長的速度，沿O—C—A的路線向點A運動；同時直線l從點B出發(fā)，以相同速度沿x軸向左平移，在平移過程中，直線l交x軸于點R，交線段BA或線段AO于點Q．當(dāng)點P到達(dá)點A時，點P和直線l都停止運動．在運動過程中，設(shè)動點P運動的時間為t秒.