2022~2023學(xué)年度第一學(xué)期期末抽測高二年級(jí)數(shù)學(xué)試題一?選擇題：本題共8小題，每小題5分，共40分.在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的.1.拋物線SKIPIF1<0的準(zhǔn)線方程是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】根據(jù)拋物線準(zhǔn)線方程的概念即可選出選項(xiàng).【詳解】解:由題知SKIPIF1<0,所以SKIPIF1<0,且拋物線開口向上,所以其準(zhǔn)線方程為:SKIPIF1<0.故選:D2.雙曲線SKIPIF1<0的漸近線方程是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】由雙曲線的標(biāo)準(zhǔn)方程可直接求得雙曲線的漸近線的方程.【詳解】在雙曲線SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，因此，該雙曲線的漸近線方程為SKIPIF1<0.故選：B.【點(diǎn)睛】本題考查利用雙曲線的標(biāo)準(zhǔn)方程求漸近線方程，屬于基礎(chǔ)題.3.在SKIPIF1<0軸上截距為SKIPIF1<0，傾斜角為SKIPIF1<0的直線方程為()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】根據(jù)斜截式直接整理可得.【詳解】因?yàn)閮A斜角為SKIPIF1<0，所以斜率SKIPIF1<0.由斜截式可得直線方程為：SKIPIF1<0，即SKIPIF1<0.故選：A4.中國古代數(shù)學(xué)著作《張丘建算經(jīng)》中記載：“今有馬行轉(zhuǎn)遲，次日減半，疾七日，行七百里”.意思是說有一匹馬行走的速度逐漸減慢，每天行走的里數(shù)是前一天的一半，七天一共行走了700里路，則該馬第七天走的里數(shù)為()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】根據(jù)題意可知，每天行走的里程數(shù)成等比數(shù)列，利用等比數(shù)列的前SKIPIF1<0項(xiàng)和公式即可求得結(jié)果.【詳解】由題意得，馬每天行走的里程數(shù)成等比數(shù)列，設(shè)第SKIPIF1<0天行走的里數(shù)為SKIPIF1<0，則數(shù)列SKIPIF1<0是公比為SKIPIF1<0的等比數(shù)列；由七天一共行走了700里可得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，即該馬第七天走的里數(shù)為SKIPIF1<0.故選：B5.設(shè)SKIPIF1<0，SKIPIF1<0為實(shí)數(shù)，若直線SKIPIF1<0與圓SKIPIF1<0相交，則點(diǎn)SKIPIF1<0與圓的位置關(guān)系是()A.在圓上 B.在圓外 C.在圓內(nèi) D.不能確定【答案】B【解析】【分析】根據(jù)直線與圓的位置關(guān)系，求得SKIPIF1<0滿足的關(guān)系式，結(jié)合點(diǎn)與圓位置關(guān)系的判斷方法，判斷即可.【詳解】根據(jù)題意SKIPIF1<0，即SKIPIF1<0，故點(diǎn)SKIPIF1<0在圓SKIPIF1<0外.故選：B.6.已知集合SKIPIF1<0和SKIPIF1<0分別是由數(shù)列SKIPIF1<0和SKIPIF1<0的前100項(xiàng)組成，則SKIPIF1<0中元素的和為()A.270 B.273 C.363 D.6831【答案】A【解析】【分析】先求出數(shù)列SKIPIF1<0和SKIPIF1<0的公共項(xiàng)，滿足公共項(xiàng)小于等于數(shù)列SKIPIF1<0的100項(xiàng)，求出項(xiàng)數(shù)，然后再求和.【詳解】設(shè)數(shù)列SKIPIF1<0的第SKIPIF1<0項(xiàng)與數(shù)列SKIPIF1<0的第SKIPIF1<0項(xiàng)相等，即SKIPIF1<0，所以SKIPIF1<0.又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以數(shù)列SKIPIF1<0與數(shù)列SKIPIF1<0的公共項(xiàng)構(gòu)成的數(shù)列為SKIPIF1<0.又因?yàn)镾KIPIF1<0的第100項(xiàng)為403，而SKIPIF1<0的SKIPIF1<0，所以則SKIPIF1<0中元素的和為：SKIPIF1<0.故選：A7.設(shè)SKIPIF1<0分別是橢圓SKIPIF1<0和雙曲線SKIPIF1<0的公共焦點(diǎn)，P是它們的一個(gè)公共點(diǎn)，且SKIPIF1<0，線段SKIPIF1<0的垂直平分線經(jīng)過點(diǎn)SKIPIF1<0，若SKIPIF1<0和SKIPIF1<0的離心率分別為SKIPIF1<0，則SKIPIF1<0的值為()．A.3 B.2 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】根據(jù)題意設(shè)出橢圓的長軸長以及雙曲線的實(shí)軸長，再根據(jù)橢圓和雙曲線的定義得到SKIPIF1<0的關(guān)系，由此可求解出SKIPIF1<0的值.【詳解】設(shè)橢圓的長軸長為SKIPIF1<0，雙曲線的實(shí)軸長為SKIPIF1<0，焦距長為SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0在雙曲線的左支上，如下圖所示(不妨設(shè)SKIPIF1<0在第二象限)，因?yàn)榫€段SKIPIF1<0的垂直平分線經(jīng)過點(diǎn)SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故選：B故選：C8已知SKIPIF1<0，則()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】設(shè)SKIPIF1<0，利用導(dǎo)數(shù)可得SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，從而有SKIPIF1<0，即SKIPIF1<0；令SKIPIF1<0，利用導(dǎo)數(shù)可得SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，從而有SKIPIF1<0，即SKIPIF1<0，即可得答案.【詳解】設(shè)SKIPIF1<0，則有SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；所以SKIPIF1<0，即有SKIPIF1<0，故SKIPIF1<0；令SKIPIF1<0，則SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；所以SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0，綜上所述，則有SKIPIF1<0.故選：B【點(diǎn)睛】方法點(diǎn)睛：對(duì)于比較大小的題目，常用的方法有：(1)作差法；(2)作商法；(3)利用函數(shù)的單調(diào)性進(jìn)行比較.二?多選題：本題共4小題，每小題5分，共20分.在每小題給出的選項(xiàng)中，有多項(xiàng)符合題目要求.全部選對(duì)的得5分，部分選對(duì)的得2分，有選錯(cuò)的得0分.9.已知曲線SKIPIF1<0，則下列說法正確的是()A.若SKIPIF1<0是橢圓，則其長軸長為SKIPIF1<0B.若SKIPIF1<0，則SKIPIF1<0是雙曲線C.C不可能表示一個(gè)圓D.若SKIPIF1<0，則SKIPIF1<0上的點(diǎn)到焦點(diǎn)的最短距離為SKIPIF1<0【答案】BC【解析】【分析】根據(jù)SKIPIF1<0可知若為橢圓，則焦點(diǎn)在SKIPIF1<0軸上，進(jìn)而可判斷A,進(jìn)而可判斷BC，根據(jù)橢圓的幾何性質(zhì)可判斷D.【詳解】由于SKIPIF1<0，所以SKIPIF1<0，對(duì)于A,當(dāng)SKIPIF1<0時(shí)，故SKIPIF1<0表示焦點(diǎn)在SKIPIF1<0軸上的橢圓，故橢圓的長軸長為SKIPIF1<0，故A錯(cuò)誤,對(duì)于B,當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0是雙曲線，故B正確,對(duì)于C,由于SKIPIF1<0，故C不可能表示一個(gè)圓，故C正確，對(duì)于D,SKIPIF1<0時(shí)，SKIPIF1<0，表示焦點(diǎn)在SKIPIF1<0軸上的橢圓，且此時(shí)SKIPIF1<0故橢圓上的點(diǎn)到焦點(diǎn)的最小距離為SKIPIF1<0，故D錯(cuò)誤,故選:BC10.已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，則()A.SKIPIF1<0B.SKIPIF1<0的前10項(xiàng)和為SKIPIF1<0C.SKIPIF1<0的前11項(xiàng)和為SKIPIF1<0D.SKIPIF1<0的前16項(xiàng)和為SKIPIF1<0【答案】ACD【解析】【分析】根據(jù)遞推公式得SKIPIF1<0進(jìn)而根據(jù)等差數(shù)列的求和公式即可判斷AB,根據(jù)并項(xiàng)求和可判斷C,根據(jù)正負(fù)去絕對(duì)值以及等差數(shù)列求和可判斷D.【詳解】由SKIPIF1<0得：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，兩式相減得SKIPIF1<0，故SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0也符合，故SKIPIF1<0對(duì)于A,SKIPIF1<0,故A正確，對(duì)于B，SKIPIF1<0的前10項(xiàng)和為SKIPIF1<0，故B錯(cuò)誤，對(duì)于C，SKIPIF1<0的前11項(xiàng)和為SKIPIF1<0，故C正確,對(duì)于D,當(dāng)SKIPIF1<0，解得SKIPIF1<0所以SKIPIF1<0所以SKIPIF1<0的前16項(xiàng)和為SKIPIF1<0SKIPIF1<0,故D正確，故選：ACD11.連續(xù)曲線上凹弧與凸弧的分界點(diǎn)稱為曲線的拐點(diǎn),拐點(diǎn)在統(tǒng)計(jì)學(xué)?物理學(xué)?經(jīng)濟(jì)學(xué)等領(lǐng)域都有重要應(yīng)用.若SKIPIF1<0的圖象是一條連續(xù)不斷的曲線,SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0都存在,且SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0也都存在.若SKIPIF1<0,使得SKIPIF1<0,且在SKIPIF1<0的左?右附近,SKIPIF1<0異號(hào),則稱點(diǎn)SKIPIF1<0為曲線SKIPIF1<0的拐點(diǎn).則以下函數(shù)具有唯一拐點(diǎn)的是()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】BCD【解析】【分析】根據(jù)拐點(diǎn)的定義及零點(diǎn)存在定理對(duì)選項(xiàng)求二階導(dǎo)函數(shù),判斷其是否有異號(hào)零點(diǎn)即可.【詳解】關(guān)于選項(xiàng)A:SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,根據(jù)拐點(diǎn)定義可知,SKIPIF1<0沒有拐點(diǎn);關(guān)于選項(xiàng)B:SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,且SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0時(shí),SKIPIF1<0,故SKIPIF1<0為SKIPIF1<0的拐點(diǎn);關(guān)于選項(xiàng)C:SKIPIF1<0,SKIPIF1<0,令SKIPIF1<0,解得SKIPIF1<0,且SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0時(shí),SKIPIF1<0,故SKIPIF1<0為SKIPIF1<0的拐點(diǎn);關(guān)于選項(xiàng)D:SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,因?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,使得SKIPIF1<0成立,由于SKIPIF1<0在SKIPIF1<0是連續(xù)不斷可導(dǎo)的,所以SKIPIF1<0在SKIPIF1<0有異號(hào)函數(shù)值,故SKIPIF1<0存在拐點(diǎn).故選:BCD12.設(shè)有一組圓SKIPIF1<0，下列命題正確是()A.不論SKIPIF1<0如何變化，圓心SKIPIF1<0始終在一條直線上B.存在圓SKIPIF1<0經(jīng)過點(diǎn)SKIPIF1<0C.存在定直線始終與圓SKIPIF1<0相切D.若圓SKIPIF1<0上總存在兩點(diǎn)到原點(diǎn)的距離為SKIPIF1<0，則SKIPIF1<0【答案】ACD【解析】【分析】對(duì)于A，考查圓心SKIPIF1<0的橫縱坐標(biāo)關(guān)系即可判斷；對(duì)于B，把SKIPIF1<0，SKIPIF1<0代入圓SKIPIF1<0方程，由關(guān)于SKIPIF1<0的方程根的情況作出判斷；對(duì)于C，判斷圓心SKIPIF1<0到直線SKIPIF1<0距離與半徑的關(guān)系即可；對(duì)于D，圓SKIPIF1<0與以原點(diǎn)為圓心的單位圓相交即可判斷作答．【詳解】對(duì)于A，圓心為SKIPIF1<0，其圓心在直線SKIPIF1<0上，A正確；對(duì)于B，圓SKIPIF1<0，將SKIPIF1<0代入圓的方程可得SKIPIF1<0，化簡得SKIPIF1<0，SKIPIF1<0，方程無解，所以不存在圓SKIPIF1<0經(jīng)過點(diǎn)SKIPIF1<0，B錯(cuò)誤；對(duì)于C，存在直線SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，圓心SKIPIF1<0到直線SKIPIF1<0或SKIPIF1<0的距離SKIPIF1<0，這兩條直線始終與圓SKIPIF1<0相切，C正確，對(duì)于D，若圓SKIPIF1<0上總存在兩點(diǎn)到原點(diǎn)的距離為1，問題轉(zhuǎn)化為圓SKIPIF1<0與圓SKIPIF1<0有兩個(gè)交點(diǎn)，圓心距為SKIPIF1<0，則有SKIPIF1<0，解可得：SKIPIF1<0或SKIPIF1<0，D正確．故選：ACD．三?填空題：本題共4小題，每小題5分，共20分.13.已知直線SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的值為______.【答案】SKIPIF1<0【解析】【分析】根據(jù)兩直線平行滿足的關(guān)系即可求解.【詳解】由SKIPIF1<0可得SKIPIF1<0，得SKIPIF1<0，故答案為：SKIPIF1<014.已知等差數(shù)列SKIPIF1<0的公差SKIPIF1<0，若SKIPIF1<0成等比數(shù)列，則SKIPIF1<0的值為______.【答案】SKIPIF1<0【解析】【分析】根據(jù)等比中項(xiàng)以及等差數(shù)列基本量的計(jì)算即可化簡求解.【詳解】由SKIPIF1<0得SKIPIF1<0,所以SKIPIF1<0,故答案為：SKIPIF1<015.已知函數(shù)SKIPIF1<0,若SKIPIF1<0恒成立,則實(shí)數(shù)SKIPIF1<0的取值范圍為______.【答案】SKIPIF1<0【解析】【分析】SKIPIF1<0恒成立即SKIPIF1<0在SKIPIF1<0上恒成立,只需SKIPIF1<0即可,構(gòu)造新函數(shù)求導(dǎo)求單調(diào)性及最大值即可.【詳解】解:由題知SKIPIF1<0恒成立,即SKIPIF1<0在SKIPIF1<0上恒成立,即SKIPIF1<0在SKIPIF1<0上恒成立,即SKIPIF1<0,記SKIPIF1<0,所以SKIPIF1<0,當(dāng)SKIPIF1<0時(shí)SKIPIF1<0,SKIPIF1<0單調(diào)遞增,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0單調(diào)遞減,所以SKIPIF1<0,所以SKIPIF1<0.故答案為:SKIPIF1<016.已知拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0為SKIPIF1<0上一點(diǎn)，以線段SKIPIF1<0為直徑的圓SKIPIF1<0與SKIPIF1<0交于另外一點(diǎn)SKIPIF1<0為圓心，SKIPIF1<0為坐標(biāo)原點(diǎn).當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的長為______，點(diǎn)SKIPIF1<0到SKIPIF1<0軸的距離為______.【答案】①.SKIPIF1<0②.SKIPIF1<0【解析】【分析】易知焦點(diǎn)SKIPIF1<0，根據(jù)SKIPIF1<0在拋物線上設(shè)出坐標(biāo)，易知圓心SKIPIF1<0為SKIPIF1<0的中點(diǎn)即可求出SKIPIF1<0，由SKIPIF1<0利用斜率相等可得SKIPIF1<0，再根據(jù)直徑所對(duì)的圓周角為SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0，利用向量數(shù)量積為0可得SKIPIF1<0，聯(lián)立及可解得SKIPIF1<0，根據(jù)兩點(diǎn)間距離公式可得SKIPIF1<0，點(diǎn)SKIPIF1<0到SKIPIF1<0軸的距離為其橫坐標(biāo)的絕對(duì)值等于SKIPIF1<0.【詳解】由題意知SKIPIF1<0在拋物線上，設(shè)SKIPIF1<0，SKIPIF1<0，如下圖所示：拋物線焦點(diǎn)SKIPIF1<0，圓心SKIPIF1<0為SKIPIF1<0的中點(diǎn)，所以SKIPIF1<0由SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0，整理可得SKIPIF1<0，即SKIPIF1<0；又因?yàn)镾KIPIF1<0為直徑，且點(diǎn)SKIPIF1<0在圓SKIPIF1<0上，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，可得SKIPIF1<0，又SKIPIF1<0，即SKIPIF1<0，整理得SKIPIF1<0，聯(lián)立SKIPIF1<0可得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0(舍)所以SKIPIF1<0，因此SKIPIF1<0；點(diǎn)SKIPIF1<0到SKIPIF1<0軸的距離為SKIPIF1<0點(diǎn)橫坐標(biāo)的絕對(duì)值，即SKIPIF1<0故答案為：SKIPIF1<0，SKIPIF1<0【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：本題的關(guān)鍵在于利用幾何關(guān)系實(shí)現(xiàn)從形到數(shù)的轉(zhuǎn)化，將直線平行轉(zhuǎn)化成斜率相等，將直徑所對(duì)的圓周角為直角轉(zhuǎn)化成向量數(shù)量積為0，從而得出坐標(biāo)之間的等量關(guān)系在進(jìn)行計(jì)算求解.四?解答題：本題6小題，共70分.解答應(yīng)寫出件字說明?證明過程或演算步驟.17.在①SKIPIF1<0,②SKIPIF1<0,③SKIPIF1<0這三個(gè)條件中選擇兩個(gè),補(bǔ)充在下面問題中,并進(jìn)行解答.已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,______,______.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)設(shè)SKIPIF1<0,求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0.注:如果選擇多組條件分別解答,按第一個(gè)解答計(jì)分.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)根據(jù)SKIPIF1<0是等差數(shù)列,設(shè)出公差為SKIPIF1<0,選擇兩個(gè)選項(xiàng),將首項(xiàng)公差代入,解方程組,即可求得基本量,寫出通項(xiàng)公式;(2)根據(jù)(1)中的通項(xiàng)公式,寫出SKIPIF1<0的通項(xiàng),利用裂項(xiàng)相消即可求得前SKIPIF1<0項(xiàng)和SKIPIF1<0.【小問1詳解】由于SKIPIF1<0等差數(shù)列,設(shè)公差為SKIPIF1<0,當(dāng)選①②時(shí):SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0的通項(xiàng)公式SKIPIF1<0.選①③時(shí):SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0的通項(xiàng)公式SKIPIF1<0.選②③時(shí):SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0的通項(xiàng)公式SKIPIF1<0.【小問2詳解】由(1)知,SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0.18.已知圓SKIPIF1<0，圓SKIPIF1<0.(1)判斷SKIPIF1<0與SKIPIF1<0的位置關(guān)系；(2)若過點(diǎn)SKIPIF1<0的直線SKIPIF1<0被SKIPIF1<0、SKIPIF1<0截得的弦長之比為SKIPIF1<0，求直線SKIPIF1<0的方程.【答案】(1)外切(2)SKIPIF1<0或SKIPIF1<0【解析】【分析】(1)計(jì)算出SKIPIF1<0，利用幾何法可判斷兩圓的位置關(guān)系；(2)對(duì)直線SKIPIF1<0的斜率是否存在進(jìn)行分類討論，在直線SKIPIF1<0的斜率不存在時(shí)，直線驗(yàn)證即可；在直線SKIPIF1<0的斜率存在時(shí)，設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，利用勾股定理結(jié)合點(diǎn)到直線的距離公式可得出關(guān)于SKIPIF1<0的方程，解出SKIPIF1<0的值，即可得出直線SKIPIF1<0的方程.【小問1詳解】解：圓SKIPIF1<0的圓心為SKIPIF1<0，半徑為SKIPIF1<0，圓SKIPIF1<0的圓心為SKIPIF1<0，半徑為SKIPIF1<0.因?yàn)镾KIPIF1<0，所以圓SKIPIF1<0與圓SKIPIF1<0外切.【小問2詳解】解：當(dāng)直線SKIPIF1<0的斜率不存在時(shí)，直線SKIPIF1<0的方程為SKIPIF1<0，直線SKIPIF1<0與圓SKIPIF1<0相離，不符合題意；當(dāng)直線SKIPIF1<0的斜率存在時(shí)，設(shè)SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0，則圓心SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0，圓心SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0，所以，直線SKIPIF1<0被圓SKIPIF1<0截得的弦長為SKIPIF1<0，直線SKIPIF1<0被圓SKIPIF1<0截得的弦長為SKIPIF1<0，由題意可得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，經(jīng)檢驗(yàn)，SKIPIF1<0或SKIPIF1<0均符合題意.所以直線SKIPIF1<0的方程為SKIPIF1<0或SKIPIF1<0.19.某新建小區(qū)規(guī)劃利用一塊空地進(jìn)行配套綠化.如圖，已知空地的一邊是直路SKIPIF1<0，余下的外圍是拋物線的一段，SKIPIF1<0的中垂線恰是該拋物線的對(duì)稱軸，SKIPIF1<0是SKIPIF1<0的中點(diǎn).擬在這塊地上劃出一個(gè)等腰梯形SKIPIF1<0區(qū)域種植草坪，其中SKIPIF1<0均在該拋物線上.經(jīng)測量，直路SKIPIF1<0段長為60米，拋物線的頂點(diǎn)SKIPIF1<0到直路SKIPIF1<0的距離為40米.以SKIPIF1<0為坐標(biāo)原點(diǎn)，SKIPIF1<0所在直線為SKIPIF1<0軸建立平面直角坐標(biāo)系SKIPIF1<0.(1)求該段拋物線的方程；(2)當(dāng)SKIPIF1<0長為多少米時(shí)，等腰梯形草坪SKIPIF1<0面積最大？【答案】(1)SKIPIF1<0(2)20米【解析】【分析】(1)SKIPIF1<0，把SKIPIF1<0兩點(diǎn)坐標(biāo)代入求解即可；(2)SKIPIF1<0，由梯形的面積公式，可得梯形SKIPIF1<0的面積為SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，求導(dǎo)可知當(dāng)SKIPIF1<0時(shí)，該函數(shù)SKIPIF1<0有唯一的極大值點(diǎn)，則改點(diǎn)也是函數(shù)的最大值點(diǎn)，即可求解.【小問1詳解】設(shè)該拋物線的方程為SKIPIF1<0，由條件知，SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，故該段拋物線的方程為SKIPIF1<0.【小問2詳解】由(1)可設(shè)SKIPIF1<0，所以梯形SKIPIF1<0的面積SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上是增函數(shù)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上是減函數(shù).所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得極大值，也是最大值.故當(dāng)SKIPIF1<0長為20米時(shí)，等腰梯形草坪SKIPIF1<0的面積最大.20.已知曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與SKIPIF1<0軸的交點(diǎn)為SKIPIF1<0，且SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式；(2)設(shè)SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，求使得SKIPIF1<0成立的正整數(shù)SKIPIF1<0的最小值.【答案】(1)SKIPIF1<0(2)8【解析】【分析】(1)根據(jù)切線方程的求解得切線方程為SKIPIF1<0，SKIPIF1<0得SKIPIF1<0，即可判斷為等比數(shù)列，進(jìn)而進(jìn)行求解，(2)根據(jù)錯(cuò)位相減法求解SKIPIF1<0，即可根據(jù)SKIPIF1<0的單調(diào)性求解.【小問1詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以曲線SKIPIF1<0上點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0.令SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0是以SKIPIF1<0為首項(xiàng)，SKIPIF1<0為公比的等比數(shù)列.故SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0.【小問2詳解】由(1)知，SKIPIF1<0，所以SKIPIF1<0，兩式相減得，SKIPIF1<0，所以SKIPIF1<0.因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以使得SKIPIF1<0成立的正整數(shù)SKIPIF1<0的最小值為8.21.已知雙曲線SKIPIF1<0的左?右焦點(diǎn)分別為SKIPIF1<0，且SKIPIF1<0，過SKIPIF1<0的直線SKIPIF1<0與SKIPIF1<0的左支交于SKIPIF1<0兩點(diǎn)，當(dāng)直線SKIPIF1<0垂直于SKIPIF1<0軸時(shí)，SKIPIF1<0.(1)求SKIPIF1<0的標(biāo)準(zhǔn)方程；(2)設(shè)SKIPIF1<0為坐標(biāo)原點(diǎn)，線段SKIPIF1<0的中點(diǎn)為SKIPIF1<0，射線SKIPIF1<0交直線SKIPIF1<0于點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0在射線SKIPIF1<0上，且SKIPIF1<0，設(shè)直線SKIPIF1<0的斜率分別為SKIPIF1<0，求SKIPIF1<0的值.【答案】(1)SKIPIF1<0(2)1【解析】【分析】(1)根據(jù)題意列出關(guān)于SKIPIF1<0的方程，解出即可得結(jié)果；(2)設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立直線與雙曲線的方程結(jié)合韋達(dá)定理求出SKIPIF1<0點(diǎn)坐標(biāo)，根據(jù)題意得出SKIPIF1<0，SKIPIF1<0，由斜率計(jì)算公式即可得結(jié)果.【小問1詳解】將SKIPIF1<0代入雙曲線可得SKIPIF1<0，由條件知，SKIPIF1<0解得SKIPIF1<0.所以SKIPIF1<0的標(biāo)準(zhǔn)方程為SKIPIF1<0.【小問2詳解】設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立SKIPIF1<0消去SKIPIF1<0并整理得，SKIPIF1<0，則SKIPIF1<0設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0.所以直線SKIPIF1<0的方程為SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.所以SKIPIF1<0.【點(diǎn)睛】方法點(diǎn)睛：利用韋達(dá)定理法解決直線與圓錐曲線相交問題的基本步驟如下：(1)設(shè)直線方程，設(shè)交點(diǎn)坐標(biāo)為SKIPIF1<0；(2)聯(lián)立直線與圓錐曲線的方程，得