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93.(08青海省卷28題)王亮同學(xué)善于改進(jìn)學(xué)習(xí)方法，他發(fā)現(xiàn)對解題過程進(jìn)行回顧反思，效果會更好．某一天他利用30分鐘時間進(jìn)行自主學(xué)習(xí)．假設(shè)他用于解題的時間(單位：分鐘)與學(xué)習(xí)收益量的關(guān)系如圖甲所示，用于回顧反思的時間(單位：分鐘)與學(xué)習(xí)收益量的關(guān)系如圖乙所示(其中是拋物線的一部分，為拋物線的頂點)，且用于回顧反思的時間不超過用于解題的時間．

(1)求王亮解題的學(xué)習(xí)收益量與用于解題的時間之間的函數(shù)關(guān)系式，并寫出自變量的取值范圍；

(2)求王亮回顧反思的學(xué)習(xí)收益量與用于回顧反思的時間之間的函數(shù)關(guān)系式；

(3)王亮如何分配解題和回顧反思的時間，才能使這30分鐘的學(xué)習(xí)收益總量最大？

(學(xué)習(xí)收益總量解題的學(xué)習(xí)收益量回顧反思的學(xué)習(xí)收益量)

(08青海省卷28題解析)解：(1)設(shè)，

把代入，得．

．······································································································ (1分)

自變量的取值范圍是：．···························································· (2分)

(2)當(dāng)時，

設(shè)，···················································································· (3分)

把代入，得，．

．································································· (5分)

當(dāng)時，

············································································································· (6分)

即．

(3)設(shè)王亮用于回顧反思的時間為分鐘，學(xué)習(xí)效益總量為，

則他用于解題的時間為分鐘．

當(dāng)時，

．························· (7分)

當(dāng)時，．············································································ (8分)

當(dāng)時，

．····································································· (9分)

隨的增大而減小，

當(dāng)時，．

綜合所述，當(dāng)時，，此時．································ (10分)

即王亮用于解題的時間為26分鐘，用于回顧反思的時間為4分鐘時，學(xué)習(xí)收益總量最大．

······················································································································· (11分)

92.(08青海西寧28題)如圖14，已知半徑為1的與軸交于兩點，為的切線，切點為，圓心的坐標(biāo)為，二次函數(shù)的圖象經(jīng)過兩點．

(1)求二次函數(shù)的解析式；

(2)求切線的函數(shù)解析式；

(3)線段上是否存在一點，使得以為頂點的三角形與相似．若存在，請求出所有符合條件的點的坐標(biāo)；若不存在，請說明理由．

(08青海西寧28題解析)解：(1)圓心的坐標(biāo)為，半徑為1，，……1分

二次函數(shù)的圖象經(jīng)過點，

可得方程組················································································ 2分

解得：二次函數(shù)解析式為············································· 3分

(2)過點作軸，垂足為．······························································· 4分

是的切線，為切點，(圓的切線垂直于經(jīng)過切點的半徑)．

在中，

為銳角，···························· 5分

，

在中，．

．

點坐標(biāo)為························································································· 6分

設(shè)切線的函數(shù)解析式為，由題意可知，······ 7分

切線的函數(shù)解析式為···································································· 8分

(3)存在．············································································································ 9分

①過點作軸，與交于點．可得(兩角對應(yīng)相等兩三角形相似)

，············································ 10分

②過點作，垂足為，過點作，垂足為．

可得(兩角對應(yīng)相等兩三角開相似)

在中，，，

在中，，

，······································· 11分

符合條件的點坐標(biāo)有，······················································ 12分

91.(08內(nèi)蒙古赤峰25題)(本題滿分14分)

在平面直角坐標(biāo)系中給定以下五個點．

(1)請從五點中任選三點，求一條以平行于軸的直線為對稱軸的拋物線的解析式；

(2)求該拋物線的頂點坐標(biāo)和對稱軸，并畫出草圖；

(3)已知點在拋物線的對稱軸上，直線過點且垂直于對稱軸．驗證：以為圓心，為半徑的圓與直線相切．請你進(jìn)一步驗證，以拋物線上的點為圓心為半徑的圓也與直線相切．由此你能猜想到怎樣的結(jié)論．

(08內(nèi)蒙古赤峰25題解析)25．解：(1)設(shè)拋物線的解析式為，

且過點，

由在H ．

則．········································································································ (2分)

得方程組，

解得．

拋物線的解析式為················ (4分)

(2)由············· (6分)

得頂點坐標(biāo)為，對稱軸為．·········· (8分)

(3)①連結(jié)，過點作直線的垂線，垂足為，

則．

在中，，，

，

，

以點為圓心，為半徑的與直線相切．····························· (10分)

②連結(jié)過點作直線的垂線，垂足為．過點作垂足為，

則．

在中，，．

．

以點為圓心為半徑的與直線相切．································ (12分)

③以拋物線上任意一點為圓心，以為半徑的圓與直線相切．····· (14分)

90.(08遼寧12市26題)(本題14分)26．如圖16，在平面直角坐標(biāo)系中，直線與軸交于點，與軸交于點，拋物線經(jīng)過三點．

(1)求過三點拋物線的解析式并求出頂點的坐標(biāo)；

(2)在拋物線上是否存在點，使為直角三角形，若存在，直接寫出點坐標(biāo)；若不存在，請說明理由；

(3)試探究在直線上是否存在一點，使得的周長最小，若存在，求出點的坐標(biāo)；若不存在，請說明理由．

(08遼寧12市26題解析)

解：(1)直線與軸交于點，與軸交于點．

，························································································· 1分

點都在拋物線上，

拋物線的解析式為························································ 3分

頂點······························································································· 4分

(2)存在··············································································································· 5分

············································································································· 7分

············································································································ 9分

(3)存在·············································································································· 10分

理由：

解法一：

延長到點，使，連接交直線于點，則點就是所求的點．

　　　　　　　　　　　 ····················································································· 11分

過點作于點．

點在拋物線上，

在中，，

，，

在中，，

，，··············································· 12分

設(shè)直線的解析式為

　 解得

································································································ 13分

　 解得　

在直線上存在點，使得的周長最小，此時．··· 14分

解法二：

過點作的垂線交軸于點，則點為點關(guān)于直線的對稱點．連接交于點，則點即為所求．················································································ 11分

過點作軸于點，則，．

，

同方法一可求得．

在中，，，可求得，

為線段的垂直平分線，可證得為等邊三角形，

垂直平分．

即點為點關(guān)于的對稱點．············································· 12分

設(shè)直線的解析式為，由題意得

　 解得

································································································ 13分

　 解得　

在直線上存在點，使得的周長最小，此時．··· 14分

89.(08遼寧沈陽26題)(本題14分)26．如圖所示，在平面直角坐標(biāo)系中，矩形的邊在軸的負(fù)半軸上，邊在軸的正半軸上，且，，矩形繞點按順時針方向旋轉(zhuǎn)后得到矩形．點的對應(yīng)點為點，點的對應(yīng)點為點，點的對應(yīng)點為點，拋物線過點．

(1)判斷點是否在軸上，并說明理由；

(2)求拋物線的函數(shù)表達(dá)式；

(3)在軸的上方是否存在點，點，使以點為頂點的平行四邊形的面積是矩形面積的2倍，且點在拋物線上，若存在，請求出點，點的坐標(biāo)；若不存在，請說明理由．

(08遼寧沈陽26題解析)解：(1)點在軸上　 1分

理由如下：

連接，如圖所示，在中，，，

，

由題意可知：

點在軸上，點在軸上．········································································· 3分

(2)過點作軸于點

，

在中，，

點在第一象限，

點的坐標(biāo)為·························································································· 5分

由(1)知，點在軸的正半軸上

點的坐標(biāo)為

點的坐標(biāo)為···························································································· 6分

拋物線經(jīng)過點，

由題意，將，代入中得

　 解得

所求拋物線表達(dá)式為：······················································· 9分

(3)存在符合條件的點，點．········································································· 10分

理由如下：矩形的面積

以為頂點的平行四邊形面積為．

由題意可知為此平行四邊形一邊，

又

邊上的高為2··································································································· 11分

依題意設(shè)點的坐標(biāo)為

點在拋物線上

解得，，

，

以為頂點的四邊形是平行四邊形，

，，

當(dāng)點的坐標(biāo)為時，

點的坐標(biāo)分別為，；

當(dāng)點的坐標(biāo)為時，

點的坐標(biāo)分別為，．......................................14分

88.(08廣東肇慶25題)(本小題滿分10分)

已知點A(a，)、B(2a，y)、C(3a，y)都在拋物線上.

(1)求拋物線與x軸的交點坐標(biāo)；

(2)當(dāng)a=1時，求△ABC的面積；

(3)是否存在含有、y、y，且與a無關(guān)的等式？如果存在，試給出一個，并加以證明；如果不存在，說明理由.

(08廣東肇慶25題解析)(本小題滿分10分)

解：(1)由5=0，·············································································· (1分)

得，．·················································································· (2分)

∴拋物線與x軸的交點坐標(biāo)為(0，0)、(，0)．······································· (3分)

(2)當(dāng)a=1時，得A(1，17)、B(2，44)、C(3，81)，······························ (4分)

分別過點A、B、C作x軸的垂線，垂足分別為D、E、F，則有

=S - - ···················································· (5分)

　　　　 =--···································· (6分)

=5(個單位面積)········································································ (7分)

(3)如：． ········································································· (8分)

事實上， =45a²+36a．　　　 　　　　　　　　　　　　

　　　　 3()=3[5×(2a)²+12×2a-(5a²+12a)] =45a²+36a．············ (9分)

∴． ···················································································· (10分)

87.(08廣東茂名25題)(本題滿分10分)

如圖，在平面直角坐標(biāo)系中，拋物線=－++經(jīng)過A(0，－4)、B(，0)、 C(，0)三點，且-=5．

(1)求、的值；(4分)

(2)在拋物線上求一點D，使得四邊形BDCE是以BC為對　　 角線的菱形；(3分)

(3)在拋物線上是否存在一點P，使得四邊形BPOH是以OB為對角線的菱形？若存在，求出點P的坐標(biāo)，并判斷這個菱形是否為正方形？若不存在，請說明理由．(3分)

(08廣東茂名25題解析)解：(1)解法一：

∵拋物線=－++經(jīng)過點A(0，－4)，

　 ∴=－4 ……1分

又由題意可知，、是方程－++=0的兩個根，

∴+=，　 =－=6··································································· 2分

由已知得(-)=25

又(-)=(+)－4=－24

∴ －24=25　　　　　　　　　　　　　 　　　　　

解得=± ··········································································································· 3分

當(dāng)=時，拋物線與軸的交點在軸的正半軸上，不合題意，舍去．

∴=－． ·········································································································· 4分

解法二：∵、是方程－++c=0的兩個根，

　即方程2－3+12=0的兩個根．

∴=，··········································································· 2分

∴－==5，

　　　　 解得 =±······························································································· 3分

　　　　 (以下與解法一相同．)　　

　　 (2)∵四邊形BDCE是以BC為對角線的菱形，根據(jù)菱形的性質(zhì)，點D必在拋物線的對稱軸上， 　　 5分

　　　　　 又∵=－－－4=－(+)+　 ································· 6分

　　　　　　 ∴拋物線的頂點(－，)即為所求的點D．······································· 7分

　　 (3)∵四邊形BPOH是以OB為對角線的菱形，點B的坐標(biāo)為(－6，0)，

根據(jù)菱形的性質(zhì)，點P必是直線=-3與

拋物線=－--4的交點， ···························································· 8分

　　　　 ∴當(dāng)=－3時，=－×(－3)－×(－3)－4=4，

　　　　 ∴在拋物線上存在一點P(－3，4)，使得四邊形BPOH為菱形． ·················· 9分

　　　　　 四邊形BPOH不能成為正方形，因為如果四邊形BPOH為正方形，點P的坐標(biāo)只能是(－3，3)，但這一點不在拋物線上．······································································································· 10分

86.(08湖北宜昌25題)如圖1，已知四邊形OABC中的三個頂點坐標(biāo)為O(0，0)，A(0，n)，C(m，0)．動點P從點O出發(fā)依次沿線段OA，AB，BC向點C移動，設(shè)移動路程為z，△OPC的面積S隨著z的變化而變化的圖象如圖2所示．m，n是常數(shù)， m＞1，n＞0．

(1)請你確定n的值和點B的坐標(biāo)；

(2)當(dāng)動點P是經(jīng)過點O，C的拋物線y＝ax+bx+c的頂點，且在雙曲線y＝上時，求這時四邊形OABC的面積．

(08湖北宜昌25題解析)解：(1) 從圖中可知，當(dāng)P從O向A運動時，△POC的面積S＝mz， z由0逐步增大到2,則S由0逐步增大到m，故OA＝2，n＝2 . (1分)

同理，AB＝1，故點B的坐標(biāo)是(1，2).(2分)

(2)解法一：

∵拋物線y＝ax+bx+c經(jīng)過點O(0，0)，C(m ，0),∴c＝0，b＝－am，(3分)

∴拋物線為y＝ax－amx，頂點坐標(biāo)為(，－am²).(4分)

如圖1，設(shè)經(jīng)過點O，C，P的拋物線為l.

當(dāng)P在OA上運動時，O,P都在y軸上，

這時P,O,C三點不可能同在一條拋物線上，

∴這時拋物線l不存在, 故不存在m的值..①

當(dāng)點P與C重合時，雙曲線y＝不可能經(jīng)過P,

故也不存在m的值.②(5分)

(說明：①②任做對一處評1分，兩處全對也只評一分)

當(dāng)P在AB上運動時，即當(dāng)0<x≤1時，y＝2，

拋物線l的頂點為P(，2).

∵P在雙曲線y＝上，可得 m＝，∵>2,與 x＝≤1不合，舍去.(6分)③

容易求得直線BC的解析式是：，(7分)

當(dāng)P在BC上運動，設(shè)P的坐標(biāo)為　 (x,y)，當(dāng)P是頂點時 x＝，

故得y＝＝，頂點P為(，)，

∵1< x＝<m，∴m>2，又∵P在雙曲線y＝上，

于是，×＝，化簡后得5m－22m+22＝0,

解得,,(8分)

與題意2<x＝<m不合,舍去.④(9分)

故由①②③④，滿足條件的只有一個值：.

這時四邊形OABC的面積＝＝.(10分)

(2)解法二：

∵拋物線y＝ax+bx+c經(jīng)過點O(0，0)，C(m ，0)

∴c＝0，b＝－am，(3分)

∴拋物線為y＝ax－amx，頂點坐標(biāo)P為(，－am²). (4分)

∵m＞1，∴＞0，且≠m，

∴P不在邊OA上且不與C重合. (5分)

∵P在雙曲線y＝上，∴×(－ am²)＝即a＝－ .

.①當(dāng)1＜m≤2時，＜≤1，如圖2,分別過B，P作x軸的垂線，

M，N為垂足，此時點P在線段AB上，且縱坐標(biāo)為2，

∴－am²＝2，即a＝－.

而a＝－ ，∴－ ＝－，m＝＞2，而1＜m≤2，不合題意，舍去.(6分)

②當(dāng)m≥2時，＞1，如圖3,分別過B，P作x軸的垂線，M，N為垂足，ON＞OM，

此時點P在線段CB上，易證Rt△BMC∽Rt△PNC，

∴BM∶PN＝MC∶NC，即:　 2∶PN＝(m－1)∶，∴PN＝(7分)

而P的縱坐標(biāo)為－ am²，∴＝－ am²，即a＝

而a＝－，∴－ ＝

化簡得：5m²－22m+22＝0.解得：m＝ ，(8分)

但m≥2，所以m＝舍去，(9分)

取m ＝ .

由以上,這時四邊形OABC的面積為：

(AB+OC) ×OA＝(1+m) ×2＝. (10分)

85.(08天津市卷26題)(本小題10分)

已知拋物線，

(Ⅰ)若，，求該拋物線與軸公共點的坐標(biāo)；

(Ⅱ)若，且當(dāng)時，拋物線與軸有且只有一個公共點，求的取值范圍；

(Ⅲ)若，且時，對應(yīng)的；時，對應(yīng)的，試判斷當(dāng)時，拋物線與軸是否有公共點？若有，請證明你的結(jié)論；若沒有，闡述理由．

(08天津市卷26題解析)解(Ⅰ)當(dāng)，時，拋物線為，

方程的兩個根為，．

∴該拋物線與軸公共點的坐標(biāo)是和．　 ················································ 2分

(Ⅱ)當(dāng)時，拋物線為，且與軸有公共點．

對于方程，判別式≥0，有≤． ········································ 3分

①當(dāng)時，由方程，解得．

此時拋物線為與軸只有一個公共點．································· 4分

②當(dāng)時，

時，，

時，．

由已知時，該拋物線與軸有且只有一個公共點，考慮其對稱軸為，

應(yīng)有　 即

解得．

綜上，或．　　 ················································································ 6分

(Ⅲ)對于二次函數(shù)，

由已知時，；時，，

又，∴．

于是．而，∴，即．

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

∵關(guān)于的一元二次方程的判別式

，　

∴拋物線與軸有兩個公共點，頂點在軸下方．····························· 8分

又該拋物線的對稱軸，

由，，，

得，

∴．

又由已知時，；時，，觀察圖象，

可知在范圍內(nèi)，該拋物線與軸有兩個公共點． ············································ 10分

83.(08廣東東莞22題)(本題滿分9分)將兩塊大小一樣含30°角的直角三角板，疊放在一起，使得它們的斜邊

AB重合，直角邊不重合，已知AB=8，BC=AD=4，AC與BD相交于點E，連結(jié)CD．

(1)填空：如圖9，AC=　　　　 ，BD=　　　　 ；四邊形ABCD是　　　 梯形.

(2)請寫出圖9中所有的相似三角形(不含全等三角形).

(3)如圖10，若以AB所在直線為軸，過點A垂直于AB的直線為軸建立如圖10的平面直角坐標(biāo)系，保持ΔABD不動，將ΔABC向軸的正方向平移到ΔFGH的位置，F(xiàn)H與BD相交于點P，設(shè)AF=t，ΔFBP面積為S，求S與t之間的函數(shù)關(guān)系式，并寫出t的取值值范圍.

(08廣東東莞22題解析)解：(1)，，…………………………1分

等腰；…………………………2分

　 (2)共有9對相似三角形.(寫對3－5對得1分，寫對6－8對得2分，寫對9對得3分)

　 �、佟鱀CE、△ABE與△ACD或△BDC兩兩相似，分別是：△DCE∽△ABE，△DCE∽△ACD，△DCE∽△BDC，△ABE∽△ACD，△ABE∽△BDC；(有5對)

②△ABD∽△EAD，△ABD∽△EBC；(有2對)

③△BAC∽△EAD，△BAC∽△EBC；(有2對)

所以，一共有9對相似三角形.………………………………5分

(3)由題意知，F(xiàn)P∥AE，

　　 ∴ ∠1＝∠PFB，

又∵ ∠1＝∠2＝30°，

　 ∴ ∠PFB＝∠2＝30°,

∴ FP＝BP.…………………………6分

過點P作PK⊥FB于點K，則.

∵ AF＝t，AB＝8，

∴ FB＝8－t，.

在Rt△BPK中，. ……………………7分

∴ △FBP的面積，

∴ S與t之間的函數(shù)關(guān)系式為：

　　　 ，或. …………………………………8分

t的取值范圍為：. …………………………………………………………9分

84(08甘肅蘭州28題)(本題滿分12分)如圖19-1，是一張放在平面直角坐標(biāo)系中的矩形紙片，為原點，點在軸的正半軸上，點在軸的正半軸上，，．

(1)在邊上取一點，將紙片沿翻折，使點落在邊上的點處，求兩點的坐標(biāo)；

(2)如圖19-2，若上有一動點(不與重合)自點沿方向向點勻速運動，運動的速度為每秒1個單位長度，設(shè)運動的時間為秒()，過點作的平行線交于點，過點作的平行線交于點．求四邊形的面積與時間之間的函數(shù)關(guān)系式；當(dāng)取何值時，有最大值？最大值是多少？

(3)在(2)的條件下，當(dāng)為何值時，以為頂點的三角形為等腰三角形，并求出相應(yīng)的時刻點的坐標(biāo)．

(08甘肅蘭州28題解析)(本題滿分12分)

解：(1)依題意可知，折痕是四邊形的對稱軸，

在中，，．

．．

點坐標(biāo)為(2，4)．································································································ 2分

在中，，　 又．

�。�　 解得：．

點坐標(biāo)為···································································································· 3分

(2)如圖①，．

，又知，，

， 又．

而顯然四邊形為矩形．

·························································· 5分

，又

當(dāng)時，有最大值．········································································ 6分

(3)(i)若以為等腰三角形的底，則(如圖①)

在中，，，為的中點，

．

又，為的中點．

過點作，垂足為，則是的中位線，

，，

當(dāng)時，，為等腰三角形．

此時點坐標(biāo)為．··························································································· 8分

(ii)若以為等腰三角形的腰，則(如圖②)

在中，．

過點作，垂足為．

，．

．

，．

，，

當(dāng)時，()，此時點坐標(biāo)為．·························· 11分

綜合(i)(ii)可知，或時，以為頂點的三角形為等腰三角形，相應(yīng)點的坐標(biāo)為或．·········································································································· 12分