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專題15直線與拋物線的位置關(guān)系限時(shí):120分鐘滿分:150分一、單選題:本大題共8小題,每個(gè)小題5分,共40分.在每小題給出的選項(xiàng)中,只有一項(xiàng)是符合題目要求的.1.已知拋物線SKIPIF1<0,過點(diǎn)SKIPIF1<0的直線l與拋物線C交于A,B兩點(diǎn),若SKIPIF1<0,則直線l的斜率是(

)A.SKIPIF1<0 B.3 C.SKIPIF1<0 D.6【解析】設(shè)SKIPIF1<0,SKIPIF1<0,因?yàn)镾KIPIF1<0,所以點(diǎn)P是線段AB的中點(diǎn),則SKIPIF1<0.因?yàn)锳,B都是拋物線C上的點(diǎn),所以SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,則SKIPIF1<0.故選:A2.已知拋物線C的方程為SKIPIF1<0,過點(diǎn)SKIPIF1<0和點(diǎn)SKIPIF1<0的直線l與拋物線C沒有公共點(diǎn),則實(shí)數(shù)t的取值范圍是()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【解析】當(dāng)SKIPIF1<0時(shí),直線SKIPIF1<0,與拋物線SKIPIF1<0有交點(diǎn),所以SKIPIF1<0,設(shè)直線SKIPIF1<0的方程為SKIPIF1<0,聯(lián)立直線與拋物線方程,得SKIPIF1<0,消元整理,得SKIPIF1<0,由于直線與拋物線無公共點(diǎn),即方程SKIPIF1<0無解,故有SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0.故選:A3.過拋物線SKIPIF1<0上定點(diǎn)SKIPIF1<0作圓SKIPIF1<0的兩條切線,分別交拋物線SKIPIF1<0于另外兩點(diǎn)SKIPIF1<0、SKIPIF1<0,則直線SKIPIF1<0的方程為(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【解析】圓SKIPIF1<0的圓心為SKIPIF1<0,半徑為SKIPIF1<0,易知SKIPIF1<0軸,所以,直線SKIPIF1<0、SKIPIF1<0的斜率必然存在,設(shè)過點(diǎn)SKIPIF1<0且與圓SKIPIF1<0相切的直線的方程為SKIPIF1<0,即SKIPIF1<0,由題意可得SKIPIF1<0,解得SKIPIF1<0,設(shè)點(diǎn)SKIPIF1<0、SKIPIF1<0,不妨設(shè)直線SKIPIF1<0、SKIPIF1<0的斜率分別為SKIPIF1<0、SKIPIF1<0,則SKIPIF1<0,可得SKIPIF1<0,同理SKIPIF1<0,可得SKIPIF1<0,直線SKIPIF1<0的斜率為SKIPIF1<0,易知點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0,所以,直線SKIPIF1<0的方程為SKIPIF1<0,即SKIPIF1<0.故選:B.4.已知拋物線SKIPIF1<0的焦點(diǎn)為F,直線l過點(diǎn)F且與C交于M,N兩點(diǎn),若SKIPIF1<0,則SKIPIF1<0的面積為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【解析】已知拋物線SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0焦點(diǎn)SKIPIF1<0由拋物線的定義可知SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則直線SKIPIF1<0,聯(lián)立SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,故選:C.5.已知點(diǎn)SKIPIF1<0為拋物線SKIPIF1<0:SKIPIF1<0的焦點(diǎn),過點(diǎn)F且傾斜角為60°的直線交拋物線SKIPIF1<0于A,B兩點(diǎn),若SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.1 C.SKIPIF1<0 D.2【解析】由題意知SKIPIF1<0的方程為SKIPIF1<0,代入SKIPIF1<0的方程,得SKIPIF1<0,設(shè)SKIPIF1<0,則SKIPIF1<0;因?yàn)镾KIPIF1<0,且SKIPIF1<0,所以SKIPIF1<0,整理得SKIPIF1<0,所以SKIPIF1<0,結(jié)合SKIPIF1<0,解得SKIPIF1<0.故選:C6.拋物線SKIPIF1<0上一點(diǎn)SKIPIF1<0到直線SKIPIF1<0距離的最小值為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【解析】設(shè)直線SKIPIF1<0SKIPIF1<0與SKIPIF1<0相切,聯(lián)立SKIPIF1<0與SKIPIF1<0得:SKIPIF1<0,由SKIPIF1<0,得:SKIPIF1<0,則直線SKIPIF1<0為SKIPIF1<0,故SKIPIF1<0與SKIPIF1<0之間的距離即為SKIPIF1<0上一點(diǎn)SKIPIF1<0到直線SKIPIF1<0距離的最小值,由兩平行線間距離公式得:SKIPIF1<0.故選:A7.已知點(diǎn)SKIPIF1<0和拋物線SKIPIF1<0,過拋物線SKIPIF1<0的焦點(diǎn)有斜率存在且不為0的直線與SKIPIF1<0交于SKIPIF1<0,SKIPIF1<0兩點(diǎn).若SKIPIF1<0,則直線SKIPIF1<0的方程為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【解析】由SKIPIF1<0得焦點(diǎn)坐標(biāo)為SKIPIF1<0,設(shè)直線方程為SKIPIF1<0,與拋物線方程聯(lián)立SKIPIF1<0,消去y得SKIPIF1<0,設(shè)SKIPIF1<0,則SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,所以直線方程為:SKIPIF1<0,故選:A8.已知拋物線SKIPIF1<0:SKIPIF1<0和圓SKIPIF1<0:SKIPIF1<0,過SKIPIF1<0點(diǎn)作直線SKIPIF1<0與上述兩曲線自左而右依次交于點(diǎn)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的最小值為(

)A.SKIPIF1<0 B.2 C.3 D.SKIPIF1<0【解析】由拋物線SKIPIF1<0:SKIPIF1<0可知焦點(diǎn)為SKIPIF1<0,設(shè)直線SKIPIF1<0的方程為SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0,設(shè)SKIPIF1<0,則SKIPIF1<0,由拋物線的定義可知SKIPIF1<0∴SKIPIF1<0,∴SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào).故選:D二、多選題:本大題共4小題,每個(gè)小題5分,共20分.在每小題給出的選項(xiàng)中,只有一項(xiàng)或者多項(xiàng)是符合題目要求的.9.若直線SKIPIF1<0與拋物線SKIPIF1<0只有一個(gè)公共點(diǎn),則實(shí)數(shù)k的值可以為()A.SKIPIF1<0 B.0 C.8 D.-8【解析】聯(lián)立SKIPIF1<0與SKIPIF1<0得,SKIPIF1<0,若SKIPIF1<0,直線與拋物線只有一個(gè)交點(diǎn),滿足要求,若SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,綜上可知SKIPIF1<0或SKIPIF1<0.故選:AB10.在平面直角坐標(biāo)系xOy中,已知拋物線SKIPIF1<0的焦點(diǎn)為F,直線l的傾斜角為60°且經(jīng)過點(diǎn)F.若l與C相交于SKIPIF1<0兩點(diǎn),則(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.△AOB的面積為SKIPIF1<0【解析】拋物線SKIPIF1<0的焦點(diǎn)坐標(biāo)為SKIPIF1<0,所以直線SKIPIF1<0:SKIPIF1<0,則SKIPIF1<0,消去SKIPIF1<0得SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,故A錯(cuò)誤,C正確;SKIPIF1<0,故B正確;又SKIPIF1<0到直線SKIPIF1<0:SKIPIF1<0的距離SKIPIF1<0,所以SKIPIF1<0,故D錯(cuò)誤;故選:BC11.過拋物線SKIPIF1<0的焦點(diǎn)作一條直線與拋物線相交于SKIPIF1<0、SKIPIF1<0兩點(diǎn),它們的橫坐標(biāo)之和等于SKIPIF1<0,則這樣的直線方程為(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.不存在【解析】拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0,若直線SKIPIF1<0與SKIPIF1<0軸重合,則該直線與拋物線只有一個(gè)交點(diǎn),不合乎題意.設(shè)直線SKIPIF1<0的方程為SKIPIF1<0,設(shè)點(diǎn)SKIPIF1<0、SKIPIF1<0,聯(lián)立SKIPIF1<0,可得SKIPIF1<0,SKIPIF1<0,由韋達(dá)定理可得SKIPIF1<0,則SKIPIF1<0,解得SKIPIF1<0,所以,直線SKIPIF1<0的方程為SKIPIF1<0,即SKIPIF1<0.故選:BC.12.在平面直角坐標(biāo)系SKIPIF1<0中,已知SKIPIF1<0為拋物線SKIPIF1<0的焦點(diǎn),點(diǎn)SKIPIF1<0在該拋物線上且位于SKIPIF1<0軸的兩側(cè),SKIPIF1<0,則(

)A.SKIPIF1<0 B.直線SKIPIF1<0過點(diǎn)SKIPIF1<0C.SKIPIF1<0的面積最小值是SKIPIF1<0 D.SKIPIF1<0與SKIPIF1<0面積之和的最小值是SKIPIF1<0【解析】設(shè)SKIPIF1<0:SKIPIF1<0,SKIPIF1<0,消SKIPIF1<0可得SKIPIF1<0.SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,則SKIPIF1<0或SKIPIF1<0∵SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,故A錯(cuò);SKIPIF1<0:SKIPIF1<0過SKIPIF1<0,故B對(duì);設(shè)定點(diǎn)SKIPIF1<0,SKIPIF1<0SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí),取等號(hào),故C對(duì);又SKIPIF1<0,不妨設(shè)SKIPIF1<0,又SKIPIF1<0,SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí),取等號(hào),故D對(duì).故選:BCD.三、填空題:本大題共4小題,每小題5分,共20分.把答案填在答題卡中的橫線上.13.已知拋物線SKIPIF1<0的頂點(diǎn)為坐標(biāo)原點(diǎn),準(zhǔn)線為SKIPIF1<0,直線SKIPIF1<0與拋物線SKIPIF1<0交于SKIPIF1<0兩點(diǎn),若線段SKIPIF1<0的中點(diǎn)為SKIPIF1<0,則直線SKIPIF1<0的方程為.【解析】因?yàn)閽佄锞€SKIPIF1<0的頂點(diǎn)為坐標(biāo)原點(diǎn),準(zhǔn)線為SKIPIF1<0,所以易得拋物線的方程為SKIPIF1<0,設(shè)SKIPIF1<0,因?yàn)榫€段SKIPIF1<0的中點(diǎn)為SKIPIF1<0,故SKIPIF1<0,則SKIPIF1<0,由SKIPIF1<0,兩式相減得SKIPIF1<0,所以SKIPIF1<0,故直線SKIPIF1<0的方程為SKIPIF1<0,即SKIPIF1<0.

14.已知拋物線SKIPIF1<0上的兩個(gè)不同的點(diǎn)SKIPIF1<0,SKIPIF1<0的橫坐標(biāo)恰好是方程SKIPIF1<0的根,則直線SKIPIF1<0的方程為.【解析】由題意,直線SKIPIF1<0的斜率存在,設(shè)直線SKIPIF1<0的方程為SKIPIF1<0,因?yàn)辄c(diǎn)SKIPIF1<0,SKIPIF1<0的橫坐標(biāo)恰好是方程SKIPIF1<0的根,所以SKIPIF1<0,聯(lián)立SKIPIF1<0,消SKIPIF1<0得SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,經(jīng)檢驗(yàn),符合題意,所以直線SKIPIF1<0的方程為SKIPIF1<0.15.已知O為坐標(biāo)原點(diǎn),A,B為拋物線SKIPIF1<0SKIPIF1<0上異于點(diǎn)O的兩個(gè)動(dòng)點(diǎn),且SKIPIF1<0.若點(diǎn)O到直線AB的距離的最大值為8,則p的值為.【解析】由題意,直線SKIPIF1<0均有斜率且不為0.設(shè)直線SKIPIF1<0的方程為SKIPIF1<0,聯(lián)立方程SKIPIF1<0,解得點(diǎn)SKIPIF1<0,∵直線SKIPIF1<0的方程為SKIPIF1<0,∴SKIPIF1<0,∴直線SKIPIF1<0的方程為SKIPIF1<0,即SKIPIF1<0,令SKIPIF1<0得,SKIPIF1<0,∴直線SKIPIF1<0必過定點(diǎn)SKIPIF1<0,∴當(dāng)直線SKIPIF1<0垂直于SKIPIF1<0軸時(shí),點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離的最大,∴SKIPIF1<0,∴SKIPIF1<0.16.已知拋物線C:SKIPIF1<0的焦點(diǎn)為F,過F點(diǎn)傾斜角為SKIPIF1<0的直線與曲線C交于A、B兩點(diǎn)(A在B的右側(cè)),則SKIPIF1<0.【解析】SKIPIF1<0拋物線C:SKIPIF1<0,SKIPIF1<0焦點(diǎn)坐標(biāo)為SKIPIF1<0,SKIPIF1<0過點(diǎn)SKIPIF1<0的直線的傾斜角為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0直線的參數(shù)方程為SKIPIF1<0(SKIPIF1<0為參數(shù)),代入拋物線方程可得:SKIPIF1<0,解得:SKIPIF1<0,則SKIPIF1<0.四、解答題:本大題共6小題,共70分.解答應(yīng)寫出必要的文字說明、證明過程或演算步驟.17.點(diǎn)SKIPIF1<0為拋物線SKIPIF1<0上一點(diǎn),SKIPIF1<0為其焦點(diǎn),已知SKIPIF1<0.(1)求SKIPIF1<0與SKIPIF1<0的值;(2)以SKIPIF1<0點(diǎn)為切點(diǎn)作拋物線的切線,交y軸于點(diǎn)N,求SKIPIF1<0的面積.【解析】(1)由拋物線的定義可知SKIPIF1<0,即SKIPIF1<0,拋物線的方程為SKIPIF1<0.又SKIPIF1<0在拋物線上,所以SKIPIF1<0,故SKIPIF1<0,SKIPIF1<0.(2)設(shè)過M點(diǎn)的方程為SKIPIF1<0,由SKIPIF1<0,消去SKIPIF1<0得SKIPIF1<0,即SKIPIF1<0,令SKIPIF1<0,解得SKIPIF1<0,所以切線方程為SKIPIF1<0.令SKIPIF1<0,得SKIPIF1<0,即SKIPIF1<0,又SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.

18.已知直線SKIPIF1<0上有一個(gè)動(dòng)點(diǎn)Q,過Q作直線l垂直于x軸,動(dòng)點(diǎn)P在直線l上,且SKIPIF1<0SKIPIF1<0,記點(diǎn)P的軌跡為C.

(1)求曲線C的方程.(2)設(shè)直線l與x軸交于點(diǎn)A,且SKIPIF1<0.試判斷直線PB與曲線C的位置關(guān)系,并證明你的結(jié)論.【解析】(1)設(shè)P的坐標(biāo)為SKIPIF1<0,則點(diǎn)Q的坐標(biāo)為SKIPIF1<0.因?yàn)镾KIPIF1<0SKIPIF1<0,所以SKIPIF1<0.所以SKIPIF1<0.∴點(diǎn)P的軌跡方程為SKIPIF1<0SKIPIF1<0.(2)

直線PB與曲線C相切,設(shè)點(diǎn)P的坐標(biāo)為SKIPIF1<0,點(diǎn)A的坐標(biāo)為SKIPIF1<0.因?yàn)镾KIPIF1<0,所以SKIPIF1<0.所以點(diǎn)B的坐標(biāo)為SKIPIF1<0.所以直線PB的斜率為SKIPIF1<0.因?yàn)镾KIPIF1<0所以SKIPIF1<0.所以直線PB的方程為SKIPIF1<0代入SKIPIF1<0,得SKIPIF1<0.因?yàn)镾KIPIF1<0,所以直線PB與曲線C相切.19.已知點(diǎn)SKIPIF1<0在拋物線SKIPIF1<0:SKIPIF1<0上(1)求拋物線SKIPIF1<0的方程;(2)若直線SKIPIF1<0與拋物線SKIPIF1<0交于SKIPIF1<0,SKIPIF1<0兩點(diǎn),SKIPIF1<0,且SKIPIF1<0(其中SKIPIF1<0為坐標(biāo)原點(diǎn)),求SKIPIF1<0的最小值【解析】(1)將點(diǎn)SKIPIF1<0代入拋物線SKIPIF1<0:SKIPIF1<0中,可得SKIPIF1<0,得SKIPIF1<0,∴拋物線SKIPIF1<0的方程為SKIPIF1<0.(2)由(1)知,拋物線SKIPIF1<0的方程為SKIPIF1<0,顯然直線SKIPIF1<0的斜率不為0,設(shè)直線SKIPIF1<0的方程為SKIPIF1<0,SKIPIF1<0,聯(lián)立SKIPIF1<0,整理可得SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,故SKIPIF1<0.∴SKIPIF1<0,∵SKIPIF1<0,∴SKIPIF1<0,即SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,解得SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào).∴SKIPIF1<0的最小值為8.20.已知拋物線SKIPIF1<0上的點(diǎn)SKIPIF1<0到其焦點(diǎn)的距離為SKIPIF1<0.(1)求SKIPIF1<0和SKIPIF1<0的值;(2)若直線SKIPIF1<0交拋物線SKIPIF1<0于SKIPIF1<0、SKIPIF1<0兩點(diǎn),線段SKIPIF1<0的垂直平分線交拋物線SKIPIF1<0于SKIPIF1<0、SKIPIF1<0兩點(diǎn),求證:SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0四點(diǎn)共圓.【解析】(1)拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0,準(zhǔn)線方程為SKIPIF1<0,點(diǎn)SKIPIF1<0到其焦點(diǎn)的距離為SKIPIF1<0,則SKIPIF1<0,可得SKIPIF1<0,故拋物線SKIPIF1<0的方程為SKIPIF1<0.將點(diǎn)SKIPIF1<0的坐標(biāo)代入拋物線方程可得SKIPIF1<0,解得SKIPIF1<0.(2)解:由中垂線的性質(zhì)可得SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以,SKIPIF1<0,設(shè)SKIPIF1<0、SKIPIF1<0,聯(lián)立SKIPIF1<0消去SKIPIF1<0并整理,得SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0,即SKIPIF1<0,則SKIPIF1<0.設(shè)線段SKIPIF1<0的中點(diǎn)為SKIPIF1<0,則點(diǎn)SKIPIF1<0的縱坐標(biāo)為SKIPIF1<0,所以,點(diǎn)SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0,則SKIPIF1<0.SKIPIF1<0直線SKIPIF1<0為線段SKIPIF1<0的垂直平分線,所以,直線SKIPIF1<0的方程為SKIPIF1<0.設(shè)SKIPIF1<0、SKIPIF1<0,聯(lián)立SKIPIF1<0,消去SKIPIF1<0并整理得SKIPIF1<0,SKIPIF1<0,可得SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,故SKIPIF1<0.設(shè)線段SKIPIF1<0的中點(diǎn)為SKIPIF1<0,則SKIPIF1<0.SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,故SKIPIF1<0,所以,SKIPIF1<0,SKIPIF1<0,故SKIPIF1<0,故SKIPIF1<0,所以,點(diǎn)SKIPIF1<0、SKIPIF1<0都在以SKIPIF1<0為直徑的圓上,故SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0四點(diǎn)共圓.21.已知點(diǎn)A是拋物線x2=2py(p>0)上的動(dòng)點(diǎn),過點(diǎn)M(-1,2)的直線AM與拋物線交于另一點(diǎn)B.(1)當(dāng)A的坐標(biāo)為(-2,1)時(shí),求點(diǎn)B的坐標(biāo);(2)已知點(diǎn)P(0,2),若M為線段AB的中點(diǎn),求SKIPIF1<0面積的最大值.【解析】(1)當(dāng)SKIPIF1<0的坐標(biāo)為SKIPIF1<0時(shí),則SKIPIF1<0,所以SKIPIF1<0,所以拋物線的方程為:SKIPIF1<0,由題意可得直線SKIPIF1<0的方程為:SKIPIF1<0,即SKIPIF1<0,代入拋物線的方程可得SKIPIF1<0解得SKIPIF1<0(舍)或6,所以,SKIPIF1<0的坐標(biāo)為SKIPIF1<0(2)法一:設(shè)直線SKIPIF1<0的方程:SKIPIF1<0,即SKIPIF1<0,設(shè)直線SKIPIF1<0與SKIPIF1<0軸的交點(diǎn)為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,由SKIPIF1<0可得SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,因?yàn)镾KIPIF1<0為線段SKIPIF1<0的中點(diǎn),所以SKIPIF1<0令SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0則SKIPIF1<0的面積SKIPIF1<0SKIPIF1<0,把SKIPIF1<0代入上式,SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0的面積的最大值為2.(2)法二:SKIPIF1<0??傻肧KIPIF1<0,SKIPIF1<0,SKIPIF1<0,因?yàn)镾KIPIF1<0

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