新高考數(shù)學(xué)一輪復(fù)習(xí)第8章 第07講 拋物線 精講（教師版）

第07講拋物線(精講）目錄第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶第二部分：課前自我評(píng)估測(cè)試第三部分：典型例題剖析題型一：拋物線的定義及其應(yīng)用題型二：拋物線的標(biāo)準(zhǔn)方程題型三：拋物線的簡(jiǎn)單幾何性質(zhì)題型四：與拋物線有關(guān)的最值問(wèn)題角度1：利用拋物線定義求最值角度2：利用函數(shù)思想求最值第四部分：高考真題感悟第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶知識(shí)點(diǎn)一：拋物線的定義1、拋物線的定義：平面內(nèi)與一個(gè)定點(diǎn)SKIPIF1<0和一條定直線SKIPIF1<0（其中定點(diǎn)SKIPIF1<0不在定直線SKIPIF1<0上）的距離相等的點(diǎn)的軌跡叫做拋物線，定點(diǎn)SKIPIF1<0叫做拋物線的焦點(diǎn)，定直線SKIPIF1<0叫做拋物線的準(zhǔn)線．2、拋物線的數(shù)學(xué)表達(dá)式：SKIPIF1<0(SKIPIF1<0為點(diǎn)SKIPIF1<0到準(zhǔn)線SKIPIF1<0的距離).知識(shí)點(diǎn)二：拋物線的標(biāo)準(zhǔn)方程和幾何性質(zhì)標(biāo)準(zhǔn)方程SKIPIF1<0（SKIPIF1<0）SKIPIF1<0（SKIPIF1<0）SKIPIF1<0（SKIPIF1<0）SKIPIF1<0（SKIPIF1<0）圖形范圍SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0對(duì)稱(chēng)軸SKIPIF1<0軸SKIPIF1<0軸SKIPIF1<0軸SKIPIF1<0軸焦點(diǎn)坐標(biāo)SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0準(zhǔn)線方程SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0頂點(diǎn)坐標(biāo)SKIPIF1<0離心率SKIPIF1<0通徑長(zhǎng)SKIPIF1<0知識(shí)點(diǎn)三：拋物線的焦半徑公式如下：（SKIPIF1<0為焦準(zhǔn)距）（1）焦點(diǎn)SKIPIF1<0在SKIPIF1<0軸正半軸，拋物線上任意一點(diǎn)SKIPIF1<0，則SKIPIF1<0；（2）焦點(diǎn)SKIPIF1<0在SKIPIF1<0軸負(fù)半軸，拋物線上任意一點(diǎn)SKIPIF1<0，則SKIPIF1<0；（3）焦點(diǎn)SKIPIF1<0在SKIPIF1<0軸正半軸，拋物線上任意一點(diǎn)SKIPIF1<0，則SKIPIF1<0；（4）焦點(diǎn)SKIPIF1<0在SKIPIF1<0軸負(fù)半軸，拋物線上任意一點(diǎn)SKIPIF1<0，則SKIPIF1<0.第二部分：課前自我評(píng)估測(cè)試第二部分：課前自我評(píng)估測(cè)試1．（2022·湖南衡陽(yáng)·高二期末）拋物線SKIPIF1<0的焦點(diǎn)到其準(zhǔn)線的距離為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．2 D．4【答案】C解：拋物線SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，所以拋物線的焦點(diǎn)到其準(zhǔn)線的距離為SKIPIF1<0.故選：C2．（2022·北京平谷·高二期末）拋物線SKIPIF1<0的焦點(diǎn)到其準(zhǔn)線的距離是（

）A．1 B．2 C．3 D．4【答案】A解：拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，準(zhǔn)線方程為SKIPIF1<0，所以焦點(diǎn)到準(zhǔn)線的距離SKIPIF1<0；故選：A3．（2022·北京·清華附中高二階段練習(xí)）已知拋物線SKIPIF1<0：SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，點(diǎn)SKIPIF1<0在拋物線上，SKIPIF1<0，則點(diǎn)SKIPIF1<0的橫坐標(biāo)為（

）A．6 B．5 C．4 D．2【答案】C解：設(shè)點(diǎn)SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0，拋物線SKIPIF1<0的準(zhǔn)線方程為SKIPIF1<0，SKIPIF1<0點(diǎn)SKIPIF1<0在拋物線上，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．故選：C．4．（2022·四川省資中縣球溪高級(jí)中學(xué)高二階段練習(xí)（文））拋物線SKIPIF1<0的準(zhǔn)線方程是SKIPIF1<0，則實(shí)數(shù)a的值（

）A．SKIPIF1<0 B．SKIPIF1<0 C．8 D．-8【答案】A由題意得：SKIPIF1<0，解得：SKIPIF1<0.故選：A5．（2022·湖北·模擬預(yù)測(cè)）已知拋物線SKIPIF1<0，過(guò)其焦點(diǎn)F的直線l與其交與A、B兩點(diǎn)，SKIPIF1<0，其準(zhǔn)線方程為_(kāi)__________．【答案】SKIPIF1<0設(shè)線段AB中點(diǎn)為D，則F為線段BD中點(diǎn)，過(guò)A、B、D、F分別向拋物線準(zhǔn)線作垂線，垂足分別為SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，SKIPIF1<0，SKIPIF1<0由梯形中位線得SKIPIF1<0，SKIPIF1<0，∴準(zhǔn)線方程為SKIPIF1<0故答案為：SKIPIF1<0．第三部分：典型例題剖析第三部分：典型例題剖析題型一：拋物線的定義及其應(yīng)用典型例題例題1．（2022·上海普陀·二模）已知點(diǎn)SKIPIF1<0，直線SKIPIF1<0，若動(dòng)點(diǎn)SKIPIF1<0到SKIPIF1<0的距離等于SKIPIF1<0，則點(diǎn)SKIPIF1<0的軌跡是（

）A．橢圓 B．雙曲線C．拋物線 D．直線【答案】C由拋物線的定義（平面內(nèi)，到定點(diǎn)與定直線的距離相等的點(diǎn)的軌跡叫做拋物線）可知，點(diǎn)SKIPIF1<0的軌跡是拋物線.故選：C例題2．（2022·福建福州·高二期中）在平面直角坐標(biāo)系SKIPIF1<0中，動(dòng)點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離比它到定點(diǎn)SKIPIF1<0的距離小1，則SKIPIF1<0的軌跡方程為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D由題意知?jiǎng)狱c(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離與定點(diǎn)SKIPIF1<0的距離相等，由拋物線的定義知，P的軌跡是以SKIPIF1<0為焦點(diǎn)，SKIPIF1<0為準(zhǔn)線的拋物線，所以SKIPIF1<0，軌跡方程為SKIPIF1<0，故選：D例題3．（2022·全國(guó)·高三專(zhuān)題練習(xí)）動(dòng)點(diǎn)SKIPIF1<0到y(tǒng)軸的距離比它到定點(diǎn)SKIPIF1<0的距離小2，求動(dòng)點(diǎn)SKIPIF1<0的軌跡方程．【答案】SKIPIF1<0或SKIPIF1<0．解：∵動(dòng)點(diǎn)M到y(tǒng)軸的距離比它到定點(diǎn)SKIPIF1<0的距離小2，∴動(dòng)點(diǎn)M到定點(diǎn)SKIPIF1<0的距離與它到定直線SKIPIF1<0的距離相等．∴動(dòng)點(diǎn)M到軌跡是以SKIPIF1<0為焦點(diǎn)，SKIPIF1<0為準(zhǔn)線的拋物線，且SKIPIF1<0．∴拋物線的方程為SKIPIF1<0，又∵x軸上點(diǎn)SKIPIF1<0左側(cè)的點(diǎn)到y(tǒng)軸的距離比它到SKIPIF1<0點(diǎn)的距離小2，∴M點(diǎn)的軌跡方程為SKIPIF1<0②．綜上，得動(dòng)點(diǎn)M的軌跡方程為SKIPIF1<0或SKIPIF1<0．同類(lèi)題型歸類(lèi)練1．（2022·山東·青島二中高二階段練習(xí)）已知?jiǎng)訄AM與直線y=2相切，且與定圓SKIPIF1<0外切，則動(dòng)圓圓心M的軌跡方程為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A設(shè)動(dòng)圓圓心為M(x，y)，半徑為r，由題意可得M到C(0，－3)的距離與到直線y＝3的距離相等,由拋物線的定義可知，動(dòng)圓圓心的軌跡是以C(0，-3)為焦點(diǎn)，以y＝3為準(zhǔn)線的一條拋物線，所以SKIPIF1<0，其方程為SKIPIF1<0，故選：A2．（2022·江蘇·高二）與點(diǎn)SKIPIF1<0和直線SKIPIF1<0的距離相等的點(diǎn)的軌跡方程是______．【答案】SKIPIF1<0解：由拋物線的定義可得平面內(nèi)與點(diǎn)SKIPIF1<0和直線SKIPIF1<0的距離相等的點(diǎn)的軌跡為拋物線，且SKIPIF1<0為焦點(diǎn)，直線SKIPIF1<0為準(zhǔn)線，設(shè)拋物線的方程為SKIPIF1<0，可知SKIPIF1<0，解得SKIPIF1<0，所以該拋物線方程是SKIPIF1<0，故答案為：SKIPIF1<03．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知?jiǎng)狱c(diǎn)SKIPIF1<0的坐標(biāo)滿足SKIPIF1<0，則動(dòng)點(diǎn)SKIPIF1<0的軌跡方程為_(kāi)____________．【答案】SKIPIF1<0設(shè)SKIPIF1<0直線SKIPIF1<0SKIPIF1<0，則動(dòng)點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離為SKIPIF1<0，動(dòng)點(diǎn)SKIPIF1<0到直線SKIPIF1<0SKIPIF1<0的距離為SKIPIF1<0，又因?yàn)镾KIPIF1<0SKIPIF1<0，所以動(dòng)點(diǎn)M的軌跡是以SKIPIF1<0為焦點(diǎn)，SKIPIF1<0為準(zhǔn)線的拋物線，其軌跡方程為SKIPIF1<0．故答案為：SKIPIF1<0題型二：拋物線的標(biāo)準(zhǔn)方程典型例題例題1．（2022·云南曲靖·高二期末）過(guò)拋物線SKIPIF1<0的焦點(diǎn)SKIPIF1<0的直線交拋物線于點(diǎn)SKIPIF1<0，SKIPIF1<0，交其準(zhǔn)線于點(diǎn)SKIPIF1<0，若SKIPIF1<0，則此拋物線方程為_(kāi)_________．【答案】SKIPIF1<0如圖，作SKIPIF1<0準(zhǔn)線于SKIPIF1<0，SKIPIF1<0準(zhǔn)線于SKIPIF1<0，設(shè)SKIPIF1<0，由拋物線定義得SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，在直角三角形SKIPIF1<0中，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，從而得SKIPIF1<0，設(shè)準(zhǔn)線與x軸交于SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，因此拋物線方程為SKIPIF1<0.故答案為：SKIPIF1<0.例題2．（2022·全國(guó)·高二課時(shí)練習(xí)）求適合下列條件的拋物線的方程.(1)焦點(diǎn)為SKIPIF1<0，準(zhǔn)線方程為SKIPIF1<0；(2)頂點(diǎn)在原點(diǎn)，準(zhǔn)線方程為SKIPIF1<0；(3)頂點(diǎn)在原點(diǎn)，以SKIPIF1<0軸為對(duì)稱(chēng)軸，過(guò)點(diǎn)SKIPIF1<0．【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0(1)解：根據(jù)題意，可設(shè)所求拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，故所求拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0.(2)解：根據(jù)題意，可設(shè)所求拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，故所求拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0.(3)解：根據(jù)題意，設(shè)所求拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0，則SKIPIF1<0，得SKIPIF1<0，故所求拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0.同類(lèi)題型歸類(lèi)練1．（2022·全國(guó)·高二課時(shí)練習(xí)）已知點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離比點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離小SKIPIF1<0，求點(diǎn)SKIPIF1<0的軌跡方程．【答案】SKIPIF1<0解：由題意可知，點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離和點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離相等，故點(diǎn)SKIPIF1<0的軌跡是以點(diǎn)SKIPIF1<0為焦點(diǎn)，以直線SKIPIF1<0為準(zhǔn)線的拋物線，設(shè)點(diǎn)SKIPIF1<0的軌跡方程為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，故點(diǎn)SKIPIF1<0的軌跡方程為SKIPIF1<0.2．（2022·全國(guó)·高二課時(shí)練習(xí)）根據(jù)下列條件，求拋物線的標(biāo)準(zhǔn)方程、頂點(diǎn)坐標(biāo)和焦點(diǎn)坐標(biāo).(1)準(zhǔn)線方程為SKIPIF1<0；(2)準(zhǔn)線方程為SKIPIF1<0；(3)準(zhǔn)線方程為SKIPIF1<0．【答案】(1)拋物線方程為SKIPIF1<0，頂點(diǎn)坐標(biāo)為SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0(2)拋物線方程為SKIPIF1<0，頂點(diǎn)坐標(biāo)為SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0(3)拋物線方程為SKIPIF1<0，頂點(diǎn)坐標(biāo)為SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0(1)由題意設(shè)拋物線的方程為SKIPIF1<0，因?yàn)闇?zhǔn)線方程為SKIPIF1<0，所以SKIPIF1<0，得SKIPIF1<0，所以拋物線方程為SKIPIF1<0，頂點(diǎn)坐標(biāo)為SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0.(2)由題意設(shè)拋物線方程為SKIPIF1<0，因?yàn)闇?zhǔn)線方程為SKIPIF1<0，所以SKIPIF1<0，得SKIPIF1<0，所以拋物線方程為SKIPIF1<0，頂點(diǎn)坐標(biāo)為SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0.(3)由題意設(shè)拋物線方程為SKIPIF1<0，因?yàn)闇?zhǔn)線方程為SKIPIF1<0，所以SKIPIF1<0，得SKIPIF1<0，所以拋物線方程為SKIPIF1<0，頂點(diǎn)坐標(biāo)為SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0.題型三：拋物線的簡(jiǎn)單幾何性質(zhì)典型例題例題1．（2022·河南·駐馬店市基礎(chǔ)教學(xué)研究室高二期末（理））已知拋物線SKIPIF1<0：SKIPIF1<0，則過(guò)拋物線SKIPIF1<0的焦點(diǎn)，弦長(zhǎng)為整數(shù)且不超過(guò)2022的直線的條數(shù)是（

）A．4037 B．4044 C．2019 D．2022【答案】A∵拋物線C：SKIPIF1<0，即SKIPIF1<0，由拋物線的性質(zhì)可得，過(guò)拋物線焦點(diǎn)中，長(zhǎng)度最短的為垂直于y軸的那條弦，則過(guò)拋物線C的焦點(diǎn)，長(zhǎng)度最短的弦的長(zhǎng)為SKIPIF1<0，由拋物線的對(duì)稱(chēng)性可得，弦長(zhǎng)在5到2022之間的有共有SKIPIF1<0條，故弦長(zhǎng)為整數(shù)且不超過(guò)2022的直線的條數(shù)是SKIPIF1<0．故選：A．例題2．（多選）（2022·湖南永州·高二期末）已知拋物線SKIPIF1<0的焦點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0為SKIPIF1<0上任意一點(diǎn)，若點(diǎn)SKIPIF1<0，下列結(jié)論正確的是（

）A．SKIPIF1<0的最小值為2B．拋物線SKIPIF1<0關(guān)于SKIPIF1<0軸對(duì)稱(chēng)C．過(guò)點(diǎn)SKIPIF1<0與拋物線SKIPIF1<0有一個(gè)公共點(diǎn)的直線有且只有一條D．點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離與到焦點(diǎn)SKIPIF1<0距離之和的最小值為4【答案】CD設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，又拋物線的焦點(diǎn)為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0時(shí)，等號(hào)成立．所以SKIPIF1<0的最小值是1，A錯(cuò)；拋物線的焦點(diǎn)在SKIPIF1<0軸上，拋物線關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，B錯(cuò)；易知點(diǎn)SKIPIF1<0在拋物線的內(nèi)部（含有焦點(diǎn)的部分），因此過(guò)SKIPIF1<0與對(duì)稱(chēng)軸平行的直線與拋物線只有一個(gè)公共點(diǎn)，其他直線與拋物線都有兩個(gè)公共點(diǎn)，C正確；記拋物線的準(zhǔn)線為SKIPIF1<0，準(zhǔn)線方程為SKIPIF1<0，過(guò)SKIPIF1<0作SKIPIF1<0于SKIPIF1<0，過(guò)SKIPIF1<0作SKIPIF1<0于SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，所以當(dāng)SKIPIF1<0三點(diǎn)共線，即SKIPIF1<0與SKIPIF1<0重合時(shí)，SKIPIF1<0最小，最小值為SKIPIF1<0．D正確．故選：CD．同類(lèi)題型歸類(lèi)練1．（2022·全國(guó)·高三專(zhuān)題練習(xí)）點(diǎn)SKIPIF1<0到拋物線SKIPIF1<0的準(zhǔn)線的距離為6，那么拋物線的標(biāo)準(zhǔn)方程是（

）A．SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】D將SKIPIF1<0轉(zhuǎn)化為SKIPIF1<0,當(dāng)SKIPIF1<0時(shí)，拋物線開(kāi)口向上，準(zhǔn)線方程SKIPIF1<0，點(diǎn)SKIPIF1<0到準(zhǔn)線的距離為SKIPIF1<0，解得SKIPIF1<0，所以拋物線方程為SKIPIF1<0，即SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，拋物線開(kāi)口向下，準(zhǔn)線方程SKIPIF1<0，點(diǎn)SKIPIF1<0到準(zhǔn)線的距離為SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去），所以拋物線方程為SKIPIF1<0，即SKIPIF1<0.所以拋物線的方程為SKIPIF1<0或SKIPIF1<0故選：D2．（2022·福建·廈門(mén)一中高二階段練習(xí)）拋物線SKIPIF1<0上一點(diǎn)SKIPIF1<0到焦點(diǎn)的距離為SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的值為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A解：因?yàn)閽佄锞€SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0，所以SKIPIF1<0，拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，由拋物線的定義可知SKIPIF1<0，解得SKIPIF1<0.故選：A.3．（2022·四川·閬中中學(xué)高二階段練習(xí)（理））已知拋物線SKIPIF1<0，以SKIPIF1<0為圓心，半徑為5的圓與拋物線SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，若SKIPIF1<0，則SKIPIF1<0（

）A．4 B．8 C．10 D．16【答案】B以SKIPIF1<0為圓心，半徑為5的圓的方程為SKIPIF1<0,由拋物線SKIPIF1<0，得到拋物線關(guān)于x軸對(duì)稱(chēng)，又∵上面的圓的圓心在x軸上，∴圓的圖形也關(guān)于x軸對(duì)稱(chēng)，∴它們的交點(diǎn)A,B關(guān)于x軸對(duì)稱(chēng)，因?yàn)閨AB|=8，∴A,B點(diǎn)的縱坐標(biāo)的絕對(duì)值都是4，∵它們?cè)趻佄锞€上，于是A點(diǎn)的橫坐標(biāo)的值SKIPIF1<0,不妨設(shè)A在x軸上方，則A點(diǎn)的坐標(biāo)為SKIPIF1<0,又∵A在圓上，∴SKIPIF1<0,解得SKIPIF1<0,故選：B.題型四：與拋物線有關(guān)的最值問(wèn)題角度1：利用拋物線定義求最值典型例題例題1．（2022·新疆維吾爾自治區(qū)喀什第二中學(xué)高二期中（理））已知SKIPIF1<0，SKIPIF1<0為拋物線SKIPIF1<0的焦點(diǎn)，點(diǎn)SKIPIF1<0在拋物線上移動(dòng)，當(dāng)SKIPIF1<0取最小值時(shí)，點(diǎn)SKIPIF1<0的坐標(biāo)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D如圖所示，過(guò)SKIPIF1<0點(diǎn)作準(zhǔn)線SKIPIF1<0的垂線，垂足為SKIPIF1<0，由拋物線定義，知SKIPIF1<0SKIPIF1<0當(dāng)SKIPIF1<0在拋物線上移動(dòng)時(shí)，SKIPIF1<0的值在變化，顯然SKIPIF1<0移動(dòng)到SKIPIF1<0時(shí)，SKIPIF1<0三點(diǎn)共線，SKIPIF1<0最小，此時(shí)SKIPIF1<0，把SKIPIF1<0代入SKIPIF1<0，得SKIPIF1<0，所以當(dāng)SKIPIF1<0取最小值時(shí)，點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0．故選：D.例題2．（2022·廣西南寧·高二期末（理））已知拋物線SKIPIF1<0焦點(diǎn)的坐標(biāo)為SKIPIF1<0，SKIPIF1<0為拋物線上的任意一點(diǎn)，SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．3 B．4 C．5 D．SKIPIF1<0【答案】A因?yàn)閽佄锞€SKIPIF1<0焦點(diǎn)的坐標(biāo)為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0．記拋物線的準(zhǔn)線為l，作SKIPIF1<0于SKIPIF1<0，作SKIPIF1<0于SKIPIF1<0，則由拋物線的定義得SKIPIF1<0，當(dāng)且僅當(dāng)P為BA與拋物線的交點(diǎn)時(shí)，等號(hào)成立．故選：A.同類(lèi)題型歸類(lèi)練1．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，定點(diǎn)SKIPIF1<0，設(shè)SKIPIF1<0為拋物線上的動(dòng)點(diǎn)，SKIPIF1<0的最小值為_(kāi)_________，此時(shí)點(diǎn)SKIPIF1<0坐標(biāo)為_(kāi)_________．【答案】

3

SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0垂直于準(zhǔn)線，過(guò)SKIPIF1<0作SKIPIF1<0垂直于準(zhǔn)線，SKIPIF1<0，SKIPIF1<0取到最小值時(shí)，且為SKIPIF1<0；點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0的縱坐標(biāo)相同，可設(shè)點(diǎn)SKIPIF1<0為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，所以點(diǎn)SKIPIF1<0，SKIPIF1<0．故答案為：3；SKIPIF1<02．（2022·陜西安康·高二期末（文））已知M為拋物線SKIPIF1<0上的動(dòng)點(diǎn)，F(xiàn)為拋物線的焦點(diǎn)，SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)__________.【答案】4解：如圖所示：設(shè)點(diǎn)M在準(zhǔn)線上的射影為D，由拋物線的定義知SKIPIF1<0，∴要求SKIPIF1<0的最小值，即求SKIPIF1<0的最小值，當(dāng)D，M，P三點(diǎn)共線時(shí)，SKIPIF1<0最小，最小值為SKIPIF1<0.故答案為：43．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知點(diǎn)SKIPIF1<0在拋物線SKIPIF1<0上，點(diǎn)SKIPIF1<0在圓SKIPIF1<0上，則SKIPIF1<0長(zhǎng)度的最小值為_(kāi)__________.【答案】3因?yàn)閽佄锞€和圓都關(guān)于橫軸對(duì)稱(chēng)，所以不妨設(shè)SKIPIF1<0，設(shè)圓SKIPIF1<0的圓心坐標(biāo)為：SKIPIF1<0，半徑為1，因此SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0長(zhǎng)度的最小值為SKIPIF1<0，故答案為：SKIPIF1<04．（2022·重慶長(zhǎng)壽·高二期末）已知P為拋物線SKIPIF1<0上任意一點(diǎn)，F(xiàn)為拋物線的焦點(diǎn)，SKIPIF1<0為平面內(nèi)一定點(diǎn)，則SKIPIF1<0的最小值為_(kāi)_________．【答案】5由題意，拋物線的準(zhǔn)線為SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0向準(zhǔn)線作垂線，垂足為SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0共線時(shí)，和最?。贿^(guò)點(diǎn)SKIPIF1<0向準(zhǔn)線作垂線，垂足為SKIPIF1<0，則SKIPIF1<0，所以最小值為5.故答案為：5.5．（2022·上海市青浦高級(jí)中學(xué)高二階段練習(xí)）已知點(diǎn)SKIPIF1<0是拋物線SKIPIF1<0上的一個(gè)動(dòng)點(diǎn)，則點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離與SKIPIF1<0到SKIPIF1<0軸的距離之和的最小值為_(kāi)__________．【答案】1由拋物線SKIPIF1<0可知其焦點(diǎn)為SKIPIF1<0，由拋物線的定義可知SKIPIF1<0，故點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離與SKIPIF1<0到SKIPIF1<0軸的距離之和為SKIPIF1<0，即點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離與SKIPIF1<0到SKIPIF1<0軸的距離之和的最小值為1.故答案為：SKIPIF1<0.6．（2022·江蘇·高二）如圖所示，已知P為拋物線SKIPIF1<0上的一個(gè)動(dòng)點(diǎn)，點(diǎn)SKIPIF1<0，F(xiàn)為拋物線C的焦點(diǎn)，若SKIPIF1<0的最小值為3，則拋物線C的標(biāo)準(zhǔn)方程為_(kāi)_____.【答案】SKIPIF1<0過(guò)點(diǎn)P、Q分別作準(zhǔn)線的垂線，垂直分別為M、N，由拋物線定義可知SKIPIF1<0，當(dāng)P，M，Q三點(diǎn)共線時(shí)等號(hào)成立所以SKIPIF1<0，解得SKIPIF1<0所以拋物線C的標(biāo)準(zhǔn)方程為SKIPIF1<0.故答案為：SKIPIF1<0角度2：利用函數(shù)思想求最值典型例題例題1．（2022·四川瀘州·高二期末（文））動(dòng)點(diǎn)SKIPIF1<0在拋物線SKIPIF1<0上，則點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．12【答案】B設(shè)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最小值，最小值為SKIPIF1<0故選：B例題2．（2022·遼寧·東北育才學(xué)校模擬預(yù)測(cè)）已知拋物線SKIPIF1<0，圓SKIPIF1<0．若點(diǎn)SKIPIF1<0，SKIPIF1<0分別在SKIPIF1<0，SKIPIF1<0上運(yùn)動(dòng)，且設(shè)點(diǎn)SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A易知SKIPIF1<0即為拋物線SKIPIF1<0的焦點(diǎn)，即SKIPIF1<0，設(shè)SKIPIF1<0，∴SKIPIF1<0∴SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，上式SKIPIF1<0，取等條件：SKIPIF1<0，即SKIPIF1<0時(shí)，取得最小值SKIPIF1<0故選：A．例題3．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知拋物線SKIPIF1<0的焦點(diǎn)坐標(biāo)為SKIPIF1<0，則拋物線上的動(dòng)點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離SKIPIF1<0的最小值為（

）A．2 B．4 C．SKIPIF1<0 D．SKIPIF1<0【答案】C解：由題意，拋物線的標(biāo)準(zhǔn)方程為：SKIPIF1<0，設(shè)拋物線上的動(dòng)點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，則：SKIPIF1<0由SKIPIF1<0，所以SKIPIF1<0由SKIPIF1<0，所以SKIPIF1<0，即動(dòng)點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離SKIPIF1<0的最小值為SKIPIF1<0.故選：C同類(lèi)題型歸類(lèi)練1．（2022·內(nèi)蒙古·包鋼一中一模（文））已知圓SKIPIF1<0，點(diǎn)SKIPIF1<0在拋物線SKIPIF1<0上運(yùn)動(dòng)，過(guò)點(diǎn)SKIPIF1<0引直線SKIPIF1<0，SKIPIF1<0與圓SKIPIF1<0相切，切點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．2 C．SKIPIF1<0 D．8【答案】C圓SKIPIF1<0的方程：SKIPIF1<0，可知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故四邊形SKIPIF1<0的面積SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0取最小值時(shí)SKIPIF1<0最小，設(shè)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取最小值為SKIPIF1<0，SKIPIF1<0的最小值為SKIPIF1<0．故選：SKIPIF1<0．2．（2022·黑龍江大慶·三模（理））已知F是拋物線SKIPIF1<0的焦點(diǎn)，A為拋物線上的動(dòng)點(diǎn)，點(diǎn)SKIPIF1<0，則當(dāng)SKIPIF1<0取最大值時(shí)，SKIPIF1<0的值為_(kāi)__________.【答案】SKIPIF1<0設(shè)SKIPIF1<0，則SKIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，所以SKIPIF1<0取最大值時(shí)SKIPIF1<0，此時(shí)SKIPIF1<0.故答案為：SKIPIF1<03．（2022·全國(guó)·高二課時(shí)練習(xí)）若拋物線SKIPIF1<0上一點(diǎn)SKIPIF1<0到焦點(diǎn)的距離為6，P、Q分別為拋物線與圓SKIPIF1<0上的動(dòng)點(diǎn)，則SKIPIF1<0的最小值為_(kāi)_____．【答案】SKIPIF1<0##SKIPIF1<0由題設(shè)及拋物線定義知：SKIPIF1<0，可得SKIPIF1<0，故SKIPIF1<0，而SKIPIF1<0的圓心為SKIPIF1<0，半徑為1，所以SKIPIF1<0最小，則SKIPIF1<0共線且SKIPIF1<0，故只需SKIPIF1<0最小，令SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<0第四部分：高考真題感悟第四部分：高考真題感悟1．（2022·全國(guó)·高考真題（文））設(shè)F為拋物線SKIPIF1<0的焦點(diǎn)，點(diǎn)A在C上，點(diǎn)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（

）A．2 B．SKIPIF1<0 C．3 D．SKIPIF1<0【答案】B由題意得，SKIPIF1<0，則SKIPIF1<0，即點(diǎn)SKIPIF1<0到準(zhǔn)線SKIPIF1<0的距離為2，所以點(diǎn)SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0，不妨設(shè)點(diǎn)SKIPIF1<0在SKIPIF1<0軸上方，代入得，SKIPIF1<0，所以SKIPIF1<0.故選：B2．（2021·天津·高考真題）已知雙曲線SKIPIF1<0的右焦點(diǎn)與拋物線SKIPIF1<0的焦點(diǎn)重合，拋物線的準(zhǔn)線交雙曲線于A，B兩點(diǎn)，交雙曲線的漸近線于C、D兩點(diǎn)，若SKIPIF1<0．則雙曲線的離心率為（

）A．SKIPIF1<0