新高考數(shù)學(xué)二輪復(fù)習(xí)考點(diǎn)歸納與演練專題3-1 利用導(dǎo)數(shù)解決切線（公切線）問(wèn)題（含解析）

專題3-1利用導(dǎo)數(shù)解決切線（公切線）問(wèn)題目錄TOC\o"1-1"\h\u 1題型一：“在”型求切線 1題型二：“過(guò)”型求切線 5題型三：已知切線條數(shù)求參數(shù) 9題型四：判斷切線條數(shù) 13題型五：公切線問(wèn)題 16題型六：距離最小值 19題型七：等價(jià)轉(zhuǎn)化為距離 23 27題型一：“在”型求切線【典型例題】例題1．（2022·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0.曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C解：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以切點(diǎn)為SKIPIF1<0，切線的斜率SKIPIF1<0，所以切線方程為SKIPIF1<0，即SKIPIF1<0；故選：C例題2．（2022·四川·雅安中學(xué)高二期中（文））已知函數(shù)SKIPIF1<0在SKIPIF1<0上滿足SKIPIF1<0，則曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】∵函數(shù)SKIPIF1<0在SKIPIF1<0上滿足SKIPIF1<0，用SKIPIF1<0替換SKIPIF1<0得：SKIPIF1<0，∴SKIPIF1<0∴SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0∴SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0∴曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程是：SKIPIF1<0，即SKIPIF1<0.故選：C.【提分秘籍】已知SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程步驟：①求SKIPIF1<0；②SKIPIF1<0【變式演練】1．（2022·四川省遂寧市教育局模擬預(yù)測(cè)（文））已知SKIPIF1<0滿足SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為(

)A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】已知SKIPIF1<0滿足SKIPIF1<0，∴SKIPIF1<0為奇函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因此SKIPIF1<0，則x＞0時(shí)，SKIPIF1<0，曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線斜率SKIPIF1<0，又SKIPIF1<0，∴曲線SKIPIF1<0在點(diǎn)SKIPIF1<0，即(1，0)處的切線方程為SKIPIF1<0，整理得SKIPIF1<0﹒故選：C．2．（2022·河南·模擬預(yù)測(cè)（理））已知函數(shù)SKIPIF1<0的圖象經(jīng)過(guò)坐標(biāo)原點(diǎn)，則曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因?yàn)楹瘮?shù)SKIPIF1<0的圖象經(jīng)過(guò)坐標(biāo)原點(diǎn)，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0所以SKIPIF1<0．因?yàn)镾KIPIF1<0，所以SKIPIF1<0．所以所求切線方程為SKIPIF1<0，即SKIPIF1<0．故選：A.3．（2022·廣東·佛山市南海區(qū)九江中學(xué)高二階段練習(xí)）設(shè)函數(shù)SKIPIF1<0，則曲線SKIPIF1<0在點(diǎn)（3，－6）處的切線方程為（

）A．y＝9x＋21 B．y＝－9x＋19 C．y＝9x＋19 D．y＝－9x＋21【答案】D【詳解】解：因?yàn)楹瘮?shù)SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以切線的斜率為SKIPIF1<0.所以曲線SKIPIF1<0在點(diǎn)（3，－6）處的切線方程為y＋6＝－9（x－3），即y＝－9x＋21.故選：D.4．（2022·全國(guó)·高三專題練習(xí)（文））函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】依題意，SKIPIF1<0，則SKIPIF1<0，而SKIPIF1<0，于是有SKIPIF1<0，即SKIPIF1<0，所以所求切線方程為：SKIPIF1<0.故選：A題型二：“過(guò)”型求切線【典型例題】例題1．（2022·全國(guó)·高二課時(shí)練習(xí)）過(guò)點(diǎn)SKIPIF1<0作曲線SKIPIF1<0的切線，則切線方程為A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】設(shè)切點(diǎn)為（m，m3-3m），SKIPIF1<0的導(dǎo)數(shù)為SKIPIF1<0，可得切線斜率k＝3m2-3，由點(diǎn)斜式方程可得切線方程為y﹣m3+3m＝(3m2-3)（x﹣m），代入點(diǎn)SKIPIF1<0可得﹣6﹣m3+3m＝(3m2-3)（2﹣m），解得m＝0或m＝3，當(dāng)m=0時(shí)，切線方程為SKIPIF1<0，當(dāng)m=3時(shí)，切線方程為SKIPIF1<0，故選A．例題2．（2022·內(nèi)蒙古·阿拉善盟第一中學(xué)高二期末（文））已知曲線SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0的直線SKIPIF1<0與曲線SKIPIF1<0相切于點(diǎn)SKIPIF1<0，則點(diǎn)SKIPIF1<0的橫坐標(biāo)為_(kāi)_____________.【答案】0或SKIPIF1<0或SKIPIF1<0【詳解】設(shè)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0的切線方程為SKIPIF1<0，代入點(diǎn)SKIPIF1<0的坐標(biāo)有SKIPIF1<0，整理為SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，故答案為：0或SKIPIF1<0或SKIPIF1<0.【提分秘籍】函數(shù)SKIPIF1<0圖象過(guò)點(diǎn)SKIPIF1<0處的切線方程：①設(shè)切線坐標(biāo)SKIPIF1<0，②求出切線方程為SKIPIF1<0，③代入SKIPIF1<0求得SKIPIF1<0，從而得切線方程．【變式演練】1．（2022·山西太原·高三階段練習(xí)）若過(guò)點(diǎn)SKIPIF1<0的直線與函數(shù)SKIPIF1<0的圖象相切，則所有可能的切點(diǎn)橫坐標(biāo)之和為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】因?yàn)楹瘮?shù)SKIPIF1<0，所以SKIPIF1<0，設(shè)切點(diǎn)為SKIPIF1<0，則切線方程為：SKIPIF1<0，將點(diǎn)SKIPIF1<0代入得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以切點(diǎn)橫坐標(biāo)之和為SKIPIF1<0故選：D.2．（2022·全國(guó)·高三專題練習(xí)）過(guò)點(diǎn)SKIPIF1<0作曲線SKIPIF1<0的切線，則切線方程為A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】由SKIPIF1<0，得SKIPIF1<0，設(shè)切點(diǎn)為SKIPIF1<0則SKIPIF1<0，∴切線方程為SKIPIF1<0，∵切線過(guò)點(diǎn)SKIPIF1<0，SKIPIF1<0∴?ex0＝ex0(e?x0)，解得：SKIPIF1<0．∴切線方程為SKIPIF1<0，整理得：SKIPIF1<0.故選C.．3．（2022·河南省淮陽(yáng)中學(xué)高三階段練習(xí)（文））已知SKIPIF1<0，過(guò)SKIPIF1<0作曲線SKIPIF1<0的切線，切點(diǎn)在第一象限，則切線的斜率為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】解：由SKIPIF1<0，得SKIPIF1<0，設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，則切線方程為SKIPIF1<0，把點(diǎn)SKIPIF1<0代入并整理，得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去），故切線斜率為SKIPIF1<0．故選：C．4．（2022·陜西安康·高三期末（理））曲線SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0的切線方程是___________.【答案】SKIPIF1<0【詳解】由題意可得點(diǎn)SKIPIF1<0不在曲線SKIPIF1<0上，設(shè)切點(diǎn)為SKIPIF1<0，因?yàn)镾KIPIF1<0，∴所求切線的斜率SKIPIF1<0，所以SKIPIF1<0.因?yàn)辄c(diǎn)SKIPIF1<0是切點(diǎn)，所以SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0.設(shè)SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，所以SKIPIF1<0有唯一解SKIPIF1<0，則所求切線的斜率SKIPIF1<0，故所求切線方程為SKIPIF1<0.故答案為：SKIPIF1<0.題型三：已知切線條數(shù)求參數(shù)【典型例題】例題1．（2022·河南·安陽(yáng)一中高三階段練習(xí)（理））已知函數(shù)SKIPIF1<0，若過(guò)點(diǎn)SKIPIF1<0可以作出三條直線與曲線SKIPIF1<0相切，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，利用導(dǎo)數(shù)的幾何意義求得切線斜率，由直線過(guò)SKIPIF1<0得關(guān)于SKIPIF1<0的方程，此方程有3個(gè)不等的實(shí)根，方程轉(zhuǎn)化為SKIPIF1<0，是三次方程，它有3個(gè)解，則其極大值與極小值異號(hào)，由此可得SKIPIF1<0的范圍．【詳解】設(shè)切點(diǎn)坐標(biāo)SKIPIF1<0曲線SKIPIF1<0在SKIPIF1<0處的切線斜率為SKIPIF1<0，又SKIPIF1<0切線過(guò)點(diǎn)SKIPIF1<0切線斜率為SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，∵過(guò)點(diǎn)SKIPIF1<0可作曲線SKIPIF1<0的三條切線，SKIPIF1<0方程SKIPIF1<0有3個(gè)解．令SKIPIF1<0，則SKIPIF1<0圖象與SKIPIF1<0軸有3個(gè)交點(diǎn)，SKIPIF1<0的極大值與極小值異號(hào)，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0或2，SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0及SKIPIF1<0上遞增，在SKIPIF1<0上遞減，SKIPIF1<0是極大值，SKIPIF1<0是極小值，SKIPIF1<0,即SKIPIF1<0，解得SKIPIF1<0，故選：D.例題2．（2022·全國(guó)·益陽(yáng)平高學(xué)校高二期末）若過(guò)點(diǎn)SKIPIF1<0可作曲線SKIPIF1<0三條切線，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】設(shè)切點(diǎn)為SKIPIF1<0，根據(jù)導(dǎo)數(shù)的幾何意義寫(xiě)出切線的方程，代入點(diǎn)SKIPIF1<0，轉(zhuǎn)化為方程有3個(gè)根，構(gòu)造函數(shù)SKIPIF1<0，利用導(dǎo)數(shù)可知函數(shù)的極值，根據(jù)題意列出不等式組求解即可.【詳解】設(shè)切點(diǎn)為SKIPIF1<0，由SKIPIF1<0，故切線方程為SKIPIF1<0，因?yàn)镾KIPIF1<0在切線上，所以代入切線方程得SKIPIF1<0，則關(guān)于t的方程有三個(gè)不同的實(shí)數(shù)根，令SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，所以當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0為增函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0為減函數(shù)，且SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，所以只需SKIPIF1<0，解得SKIPIF1<0故選：A【提分秘籍】過(guò)點(diǎn)SKIPIF1<0可做函數(shù)SKIPIF1<0的一條（或兩條或三條）切線問(wèn)題步驟：①設(shè)切點(diǎn)SKIPIF1<0，求斜率SKIPIF1<0②求切線SKIPIF1<0③將點(diǎn)SKIPIF1<0代入切線SKIPIF1<0方程中得SKIPIF1<0④則問(wèn)題轉(zhuǎn)化為關(guān)于SKIPIF1<0的方程SKIPIF1<0就有幾個(gè)解⑤轉(zhuǎn)化為交點(diǎn)問(wèn)題或極值問(wèn)題求解.【變式演練】1．（2022·浙江大學(xué)附屬中學(xué)高三期中）若過(guò)SKIPIF1<0可做SKIPIF1<0的兩條切線，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】設(shè)切點(diǎn)為SKIPIF1<0，切線的斜率SKIPIF1<0，則切線方程為：SKIPIF1<0，把點(diǎn)SKIPIF1<0代入可得SKIPIF1<0，化為：SKIPIF1<0，則此方程有大于0的兩個(gè)實(shí)數(shù)根．則SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，故選：A．2．（2022·遼寧·高二期末）若過(guò)點(diǎn)SKIPIF1<0可以作曲線SKIPIF1<0的兩條切線，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，由于SKIPIF1<0，因此切線方程為SKIPIF1<0，又切線過(guò)點(diǎn)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，函數(shù)定義域是SKIPIF1<0，則直線SKIPIF1<0與曲線SKIPIF1<0有兩個(gè)不同的交點(diǎn)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，SKIPIF1<0在定義域內(nèi)單調(diào)遞增，不合題意；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，結(jié)合圖像知SKIPIF1<0，即SKIPIF1<0.故選：D.3．（2022·河南·馬店第一高級(jí)中學(xué)高二期中（文））已知函數(shù)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0可作曲線SKIPIF1<0的三條切線，則實(shí)數(shù)m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】解：設(shè)切點(diǎn)為SKIPIF1<0，則SKIPIF1<0，所以切線的斜率為SKIPIF1<0，又因?yàn)榍芯€過(guò)點(diǎn)SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得極大值SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得極大小值SKIPIF1<0，因?yàn)檫^(guò)點(diǎn)SKIPIF1<0可作曲線SKIPIF1<0的三條切線，所以方程SKIPIF1<0有3個(gè)解，則SKIPIF1<0，解得SKIPIF1<0，故選：D題型四：判斷切線條數(shù)【典型例題】例題1．（2022·安徽蚌埠·模擬預(yù)測(cè)（理））已知函數(shù)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作曲線SKIPIF1<0的切線，則可作切線的最多條數(shù)是______．【答案】3【詳解】∵點(diǎn)SKIPIF1<0不在函數(shù)SKIPIF1<0的圖象上，∴點(diǎn)SKIPIF1<0不是切點(diǎn)，設(shè)切點(diǎn)為SKIPIF1<0（SKIPIF1<0），由SKIPIF1<0，可得SKIPIF1<0，則切線的斜率SKIPIF1<0，∴SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，故切線有3條.故答案為：3.【提分秘籍】過(guò)點(diǎn)SKIPIF1<0可做函數(shù)SKIPIF1<0的幾條切線問(wèn)題步驟：①設(shè)切點(diǎn)SKIPIF1<0，求斜率SKIPIF1<0②求切線SKIPIF1<0③將點(diǎn)SKIPIF1<0代入切線SKIPIF1<0方程中得SKIPIF1<0④解出SKIPIF1<0即可判斷切線為幾條.【變式演練】1．（2022·全國(guó)·模擬預(yù)測(cè)（理））過(guò)點(diǎn)SKIPIF1<0作曲線SKIPIF1<0的切線，當(dāng)SKIPIF1<0時(shí)，切線的條數(shù)是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】設(shè)切點(diǎn)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0切線斜率SKIPIF1<0，SKIPIF1<0切線方程為：SKIPIF1<0；又切線過(guò)SKIPIF1<0，SKIPIF1<0；設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0恒成立，可得SKIPIF1<0圖象如下圖所示，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0與SKIPIF1<0有三個(gè)不同的交點(diǎn)，即當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0有三個(gè)不同的解，SKIPIF1<0切線的條數(shù)為SKIPIF1<0條.故選：D.2．（2022·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，則過(guò)點(diǎn)SKIPIF1<0可作曲線SKIPIF1<0的切線的條數(shù)為（

）A．0 B．1 C．2 D．3【答案】C【詳解】解：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，設(shè)切點(diǎn)為SKIPIF1<0，所以在切點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，又SKIPIF1<0在切線上，所以SKIPIF1<0，即SKIPIF1<0，整理得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以過(guò)點(diǎn)SKIPIF1<0可作曲線SKIPIF1<0的切線的條數(shù)為2.故選：C．題型五：公切線問(wèn)題【典型例題】例題1．（2022·湖北·仙桃市田家炳實(shí)驗(yàn)高級(jí)中學(xué)高三階段練習(xí)）若直線SKIPIF1<0是曲線SKIPIF1<0與SKIPIF1<0的公切線，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】設(shè)直線SKIPIF1<0與SKIPIF1<0的圖象相切于點(diǎn)SKIPIF1<0，與SKIPIF1<0的圖象相切于點(diǎn)SKIPIF1<0，求出SKIPIF1<0，SKIPIF1<0，由點(diǎn)SKIPIF1<0、點(diǎn)SKIPIF1<0在切線上，得切線方程，進(jìn)而即得.【詳解】設(shè)直線SKIPIF1<0與SKIPIF1<0的圖象相切于點(diǎn)SKIPIF1<0，與SKIPIF1<0的圖象相切于點(diǎn)SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，由點(diǎn)SKIPIF1<0在切線上，得切線方程為SKIPIF1<0；由點(diǎn)SKIPIF1<0在切線上，得切線方程為SKIPIF1<0，故SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0.故選：B.例題2．（2022·浙江金華·高三階段練習(xí)）若直線SKIPIF1<0是曲線SKIPIF1<0和SKIPIF1<0的公切線，則實(shí)數(shù)SKIPIF1<0的值是___________．【答案】SKIPIF1<0【分析】設(shè)直線SKIPIF1<0與曲線SKIPIF1<0、SKIPIF1<0分別相切于點(diǎn)SKIPIF1<0、SKIPIF1<0，利用導(dǎo)數(shù)求出曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程，以及曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程，可得出關(guān)于SKIPIF1<0、SKIPIF1<0的方程組，解出這兩個(gè)量的值，即可求得SKIPIF1<0的值.【詳解】設(shè)直線SKIPIF1<0與曲線SKIPIF1<0、SKIPIF1<0分別相切于點(diǎn)SKIPIF1<0、SKIPIF1<0，對(duì)函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0，則SKIPIF1<0，曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0，對(duì)函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0，則SKIPIF1<0，曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0，所以，SKIPIF1<0，化簡(jiǎn)可得SKIPIF1<0.故答案為：SKIPIF1<0.【提分秘籍】SKIPIF1<0是SKIPIF1<0和SKIPIF1<0的公切線問(wèn)題：①設(shè)SKIPIF1<0與SKIPIF1<0相切的切點(diǎn)為SKIPIF1<0則，求出切線方程SKIPIF1<0②設(shè)SKIPIF1<0與SKIPIF1<0相切的切點(diǎn)為SKIPIF1<0則，求出切線方程SKIPIF1<0③聯(lián)立兩切線求解.【變式演練】1．（2022·重慶市育才中學(xué)高三階段練習(xí)）若直線SKIPIF1<0（SKIPIF1<0）為曲線SKIPIF1<0與曲線SKIPIF1<0的公切線，則l的縱截距SKIPIF1<0（

）A．0 B．1 C．e D．SKIPIF1<0【答案】D【詳解】設(shè)l與SKIPIF1<0的切點(diǎn)為SKIPIF1<0，則由SKIPIF1<0，有SKIPIF1<0.同理，設(shè)l與SKIPIF1<0的切點(diǎn)為SKIPIF1<0，由SKIPIF1<0，有SKIPIF1<0.故SKIPIF1<0解得SKIPIF1<0或SKIPIF1<0則SKIPIF1<0或SKIPIF1<0.因SKIPIF1<0，所以l為SKIPIF1<0時(shí)不成立.故SKIPIF1<0，故選：D.2．（2022·湖南·長(zhǎng)沙一中高三開(kāi)學(xué)考試）若直線l：SKIPIF1<0為曲線SKIPIF1<0與曲線SKIPIF1<0的公切線（其中SKIPIF1<0為自然對(duì)數(shù)的底數(shù)，SKIPIF1<0），則實(shí)數(shù)b=___________.【答案】SKIPIF1<0或SKIPIF1<0##SKIPIF1<0或SKIPIF1<0【詳解】根據(jù)切線方程的求解，聯(lián)立方程即可解得切點(diǎn)，進(jìn)而可求SKIPIF1<0.設(shè)SKIPIF1<0與SKIPIF1<0的切點(diǎn)為SKIPIF1<0，則由SKIPIF1<0，有SKIPIF1<0.同理，設(shè)SKIPIF1<0與SKIPIF1<0的切點(diǎn)為SKIPIF1<0，由SKIPIF1<0，有SKIPIF1<0.故SKIPIF1<0由①式兩邊同時(shí)取對(duì)數(shù)得：SKIPIF1<0，將③代入②中可得：SKIPIF1<0，進(jìn)而解得SKIPIF1<0或SKIPIF1<0.則SKIPIF1<0或SKIPIF1<0故SKIPIF1<0或SKIPIF1<0.故答案為：SKIPIF1<0或SKIPIF1<0題型六：距離最小值【典型例題】例題1．（2022·江蘇·鎮(zhèn)江市實(shí)驗(yàn)高級(jí)中學(xué)高二期中）若點(diǎn)SKIPIF1<0，SKIPIF1<0分別是函數(shù)SKIPIF1<0與SKIPIF1<0圖象上的動(dòng)點(diǎn)（其中SKIPIF1<0是自然對(duì)數(shù)的底數(shù)），則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．17【答案】A【詳解】設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0令SKIPIF1<0且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0設(shè)與SKIPIF1<0平行且與SKIPIF1<0相切的直線與SKIPIF1<0切于SKIPIF1<0SKIPIF1<0SKIPIF1<0．SKIPIF1<0SKIPIF1<0則SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0，即SKIPIF1<0，故選：A．【提分秘籍】本例中設(shè)SKIPIF1<0，SKIPIF1<0，設(shè)與SKIPIF1<0平行且與SKIPIF1<0相切的直線與SKIPIF1<0切于SKIPIF1<0，由導(dǎo)數(shù)的幾何意義可求出點(diǎn)SKIPIF1<0的坐標(biāo)，則SKIPIF1<0的最小值轉(zhuǎn)化為點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離【變式演練】1．（2022·浙江省杭州第二中學(xué)高三階段練習(xí)）已知點(diǎn)P在函數(shù)SKIPIF1<0的圖像上，點(diǎn)Q是在直線SKIPIF1<0上，記SKIPIF1<0，則（

）A．M有最小值SKIPIF1<0 B．當(dāng)M取最小值時(shí)，點(diǎn)Q的橫坐標(biāo)是SKIPIF1<0C．M有最小值SKIPIF1<0 D．當(dāng)M取最小值時(shí)，點(diǎn)Q的橫坐標(biāo)是SKIPIF1<0【答案】D【詳解】將SKIPIF1<0化為SKIPIF1<0，即直線l的斜率為SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，∴當(dāng)M最小時(shí)，點(diǎn)P的坐標(biāo)為SKIPIF1<0，此時(shí)點(diǎn)P到直線SKIPIF1<0的距離為SKIPIF1<0，所以M的最小值為SKIPIF1<0；過(guò)點(diǎn)P且垂直于SKIPIF1<0的直線方程為SKIPIF1<0，聯(lián)立SKIPIF1<0，得SKIPIF1<0，即點(diǎn)Q的橫坐標(biāo)為SKIPIF1<0.故選D2．（2022·江蘇·蘇州市蘇州高新區(qū)第一中學(xué)高二期中）直線SKIPIF1<0分別與曲線SKIPIF1<0，直線SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】由題，設(shè)SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0，直線SKIPIF1<0的傾斜角為SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0最小即SKIPIF1<0最小，即為當(dāng)過(guò)點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0平行時(shí)最小，由曲線SKIPIF1<0，得SKIPIF1<0，所以切點(diǎn)為SKIPIF1<0，可求得點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離最小值為SKIPIF1<0故SKIPIF1<0，故選：C3．（2022·全國(guó)·高二專題練習(xí)）點(diǎn)A是曲線SKIPIF1<0上任意一點(diǎn)，則點(diǎn)A到直線SKIPIF1<0的最小距離為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】不妨設(shè)SKIPIF1<0，定義域?yàn)椋篠KIPIF1<0對(duì)SKIPIF1<0求導(dǎo)可得：SKIPIF1<0令SKIPIF1<0解得：SKIPIF1<0（其中SKIPIF1<0舍去）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則此時(shí)該點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離為最小根據(jù)點(diǎn)到直線的距離公式可得：SKIPIF1<0解得：SKIPIF1<0故選：A4．（2022·四川省宜賓市第四中學(xué)校高三階段練習(xí)（文））已知點(diǎn)SKIPIF1<0是函數(shù)SKIPIF1<0圖象上的點(diǎn)，點(diǎn)SKIPIF1<0是直線SKIPIF1<0上的點(diǎn)，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】當(dāng)與直線SKIPIF1<0平行的直線與SKIPIF1<0的圖象相切時(shí)，切點(diǎn)到直線SKIPIF1<0的距離為SKIPIF1<0的最小值.SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去），又SKIPIF1<0，所以切點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離即為SKIPIF1<0的最小值，即SKIPIF1<0.故選：A.題型七：等價(jià)轉(zhuǎn)化為距離【典型例題】例題1．（2022·河南南陽(yáng)·高二階段練習(xí)（理））已知SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0表示點(diǎn)SKIPIF1<0和SKIPIF1<0之間的距離的平方；SKIPIF1<0點(diǎn)SKIPIF1<0的軌跡為SKIPIF1<0，點(diǎn)SKIPIF1<0的軌跡為SKIPIF1<0，SKIPIF1<0的最小值即為SKIPIF1<0上的點(diǎn)與SKIPIF1<0上的點(diǎn)的距離的平方的最小值；SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0與SKIPIF1<0平行的曲線SKIPIF1<0的切線方程為SKIPIF1<0且切點(diǎn)為SKIPIF1<0，SKIPIF1<0上的點(diǎn)與SKIPIF1<0上的點(diǎn)的最短距離為點(diǎn)SKIPIF1<0到SKIPIF1<0的距離，即最短距離SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0的最小值為SKIPIF1<0.故選：B.【提分秘籍】在本例中根據(jù)幾何意義可知SKIPIF1<0表示點(diǎn)SKIPIF1<0和SKIPIF1<0之間的距離的平方，根據(jù)點(diǎn)SKIPIF1<0的軌跡方程，可將問(wèn)題轉(zhuǎn)化為SKIPIF1<0上的點(diǎn)與SKIPIF1<0上的點(diǎn)的距離的平方的最小值的求解；利用導(dǎo)數(shù)可求得與SKIPIF1<0平行的曲線的切線及切點(diǎn)，可知所求最小值即為切點(diǎn)到直線SKIPIF1<0距離平方的最小值，利用點(diǎn)到直線距離公式可求得結(jié)果.【變式演練】1．（2022·全國(guó)·高三專題練習(xí)）已知實(shí)數(shù)a，b，c，d滿足：SKIPIF1<0，其中e是自然對(duì)數(shù)的底數(shù)，則SKIPIF1<0的最小值是（

）A．7 B．8 C．9 D．10【答案】B【詳解】因?yàn)閷?shí)數(shù)a，b，c，d滿足：SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0.所以點(diǎn)SKIPIF1<0在曲線SKIPIF1<0上，點(diǎn)SKIPIF1<0在SKIPIF1<0上.所以SKIPIF1<0的幾何意義就是曲線SKIPIF1<0上的任一點(diǎn)到SKIPIF1<0上的任一點(diǎn)的距離的平方.由幾何意義可知，當(dāng)SKIPIF1<0的某一條切線與SKIPIF1<0平行時(shí)，兩平行線間距離最小.設(shè)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與SKIPIF1<0平行，則有：SKIPIF1<0，解得：SKIPIF1<0，即切點(diǎn)為SKIPIF1<0.此時(shí)SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0就是兩曲線間距離的最小值，所以SKIPIF1<0的最小值為SKIPIF1<0.故選：B2．（2022·江西·金溪一中高三階段練習(xí)（理））已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．2 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0可以轉(zhuǎn)化為：SKIPIF1<0是函數(shù)SKIPIF1<0圖象上的點(diǎn)，SKIPIF1<0是函數(shù)SKIPIF1<0上的點(diǎn)，SKIPIF1<0.當(dāng)與直線SKIPIF1<0平行且與SKIPIF1<0的圖象相切時(shí)，切點(diǎn)到直線SKIPIF1<0的距離為SKIPIF1<0的最小值.令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，（舍去），又SKIPIF1<0，所以切點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離即為SKIPIF1<0的最小值.所以SKIPIF1<0，所以SKIPIF1<0.故選：B.3．（2022·全國(guó)·高三專題練習(xí)）若SKIPIF1<0，則SKIPIF1<0的最小值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】解：由已知可得SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最小值即為曲線SKIPIF1<0的點(diǎn)到直線SKIPIF1<0的距離最小值的平方，設(shè)SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，曲線SKIPIF1<0與SKIPIF1<0平行的切線相切于SKIPIF1<0，則所求距離的最小值為點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離的平方，即SKIPIF1<0.故選：D.4．（2022·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】設(shè)SKIPIF1<0,SKIPIF1<0,點(diǎn)SKIPIF1<0在函數(shù)SKIPIF1<0上，點(diǎn)SKIPIF1<0在函數(shù)SKIPIF1<0上，SKIPIF1<0表示曲線SKIPIF1<0上點(diǎn)SKIPIF1<0到直線SKIPIF1<0的點(diǎn)SKIPIF1<0距離.由SKIPIF1<0，可得SKIPIF1<0，與直線SKIPIF1<0平行的直線的斜率為SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以切點(diǎn)的坐標(biāo)為SKIPIF1<0，切點(diǎn)到直線SKIPIF1<0的距離SKIPIF1<0.SKIPIF1<0的最小值為SKIPIF1<0.故選：B5．（2022·重慶市萬(wàn)州第二高級(jí)中學(xué)高二階段練習(xí)）在平面直角坐標(biāo)系SKIPIF1<0中，已知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．9 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】由SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0的最小值轉(zhuǎn)化為：SKIPIF1<0上的點(diǎn)與SKIPIF1<0上的點(diǎn)的距離的平方的最小值，由SKIPIF1<0，得：SKIPIF1<0，與SKIPIF1<0平行的直線的斜率為1，∴SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍SKIPIF1<0，可得切點(diǎn)為SKIPIF1<0，切點(diǎn)到直線SKIPIF1<0之間的距離的平方，即為SKIPIF1<0的最小值，SKIPIF1<0的最小值為：SKIPIF1<0.故選：B.一、單選題1．（2023·江西·貴溪市實(shí)驗(yàn)中學(xué)高三階段練習(xí)（理））曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以切線方程為SKIPIF1<0，即SKIPIF1<0．故選：A．2．（2022·湖北·棗陽(yáng)一中高三期中）已知函數(shù)SKIPIF1<0的圖像在SKIPIF1<0處的切線過(guò)點(diǎn)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．2 C．3 D．4【答案】B【詳解】由SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則函數(shù)在SKIPIF1<0處的切線方程為SKIPIF1<0，將SKIPIF1<0代入切線方程可得SKIPIF1<0.故選：B3．（2022·四川綿陽(yáng)·一模（理））已知直線SKIPIF1<0：SKIPIF1<0既是曲線SKIPIF1<0的切線，又是曲線SKIPIF1<0的切線，則SKIPIF1<0（

）A．0 B．SKIPIF1<0 C．0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】D【詳解】SKIPIF1<0,SKIPIF1<0,SKIPIF1<0