人教A版高中數(shù)學(xué)(選擇性必修第一冊(cè))同步講義第31講 3.3.2拋物線的簡(jiǎn)單幾何性質(zhì)（含解析）

第06講3.3.2拋物線的簡(jiǎn)單幾何性質(zhì)課程標(biāo)準(zhǔn)學(xué)習(xí)目標(biāo)①理解與掌握拋物線的幾何性質(zhì)。②通過(guò)對(duì)拋物線幾何性質(zhì)來(lái)解決與圓錐曲線有關(guān)的點(diǎn)、線、面積、周長(zhǎng)的相關(guān)計(jì)算問(wèn)題。③會(huì)解決與拋物線有關(guān)的弦、定點(diǎn)、定值與取值范圍問(wèn)題的處理。通過(guò)本節(jié)課的學(xué)習(xí)，要求掌握拋物線的性質(zhì)，并能解決與之相關(guān)的計(jì)算與證明問(wèn)題知識(shí)點(diǎn)01：拋物線的簡(jiǎn)單幾何性質(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)02：直線與拋物線的位置關(guān)系設(shè)直線SKIPIF1<0:SKIPIF1<0,拋物線:SKIPIF1<0（SKIPIF1<0）,將直線方程與拋物線方程聯(lián)立整理成關(guān)于SKIPIF1<0的方程SKIPIF1<0(1)若SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),直線與拋物線相交,有兩個(gè)交點(diǎn);當(dāng)SKIPIF1<0時(shí),直線與拋物線相切,有一個(gè)切點(diǎn);當(dāng)SKIPIF1<0時(shí),直線與拋物線相離,沒(méi)有公共點(diǎn).(2)若SKIPIF1<0,直線與拋物線有一個(gè)交點(diǎn),此時(shí)直線平行于拋物線的對(duì)稱(chēng)軸或與對(duì)稱(chēng)軸重合.因此直線與拋物線有一個(gè)公共點(diǎn)是直線與拋物線相切的必要不充分條件.【即學(xué)即練1】（2023·全國(guó)·高三專(zhuān)題練習(xí)）直線SKIPIF1<0與拋物線SKIPIF1<0的位置關(guān)系為（）A．相交 B．相切 C．相離 D．不能確定【答案】A【詳解】直線SKIPIF1<0過(guò)定點(diǎn)SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0在拋物線SKIPIF1<0內(nèi)部，∴直線SKIPIF1<0與拋物線SKIPIF1<0相交，故選：A．知識(shí)點(diǎn)03：直線和拋物線1、拋物線的通徑(過(guò)焦點(diǎn)且垂直于軸的弦)長(zhǎng)為SKIPIF1<0.2、拋物線的焦點(diǎn)弦過(guò)拋物線SKIPIF1<0（SKIPIF1<0）的焦點(diǎn)SKIPIF1<0的一條直線與它交于兩點(diǎn)SKIPIF1<0，SKIPIF1<0，則①SKIPIF1<0，SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0.【即學(xué)即練2】（2023秋·四川成都·高二校考期末）已知拋物線SKIPIF1<0，其焦點(diǎn)SKIPIF1<0到其準(zhǔn)線的距離為SKIPIF1<0，過(guò)焦點(diǎn)SKIPIF1<0且傾斜角為SKIPIF1<0的直線SKIPIF1<0交拋物線SKIPIF1<0于SKIPIF1<0兩點(diǎn)，（1）求拋物線SKIPIF1<0的方程及其焦點(diǎn)坐標(biāo)；（2）求SKIPIF1<0.【答案】（1）SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0；（2）8.【詳解】解：（1）拋物線SKIPIF1<0的焦點(diǎn)SKIPIF1<0到其準(zhǔn)線的距離為SKIPIF1<0，得SKIPIF1<0，所以拋物線SKIPIF1<0的方程為SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0.（2）過(guò)焦點(diǎn)SKIPIF1<0且傾斜角為SKIPIF1<0的直線SKIPIF1<0的方程為SKIPIF1<0，設(shè)SKIPIF1<0，聯(lián)立方程組SKIPIF1<0消去SKIPIF1<0可得SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0.說(shuō)明：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.題型01拋物線的簡(jiǎn)單性質(zhì)【典例1】（2023春·四川廣安·高二四川省廣安友誼中學(xué)?？茧A段練習(xí)）拋物線C與拋物線SKIPIF1<0關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，則拋物線C的準(zhǔn)線方程是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】∵拋物線C與拋物線SKIPIF1<0關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，∴拋物線C的方程為SKIPIF1<0，∴拋物線C的準(zhǔn)線方程是SKIPIF1<0.故選：C.【典例2】（2023·全國(guó)·高三專(zhuān)題練習(xí)）對(duì)拋物線SKIPIF1<0，下列描述正確的是（

）A．開(kāi)口向上，焦點(diǎn)為SKIPIF1<0 B．開(kāi)口向上，焦點(diǎn)為SKIPIF1<0C．開(kāi)口向右，焦點(diǎn)為SKIPIF1<0 D．開(kāi)口向右，焦點(diǎn)為SKIPIF1<0【答案】A【詳解】由題知，該拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0，則該拋物線開(kāi)口向上，焦點(diǎn)坐標(biāo)為SKIPIF1<0.故選：A.【典例3】（2023秋·高二課時(shí)練習(xí)）根據(jù)下列條件寫(xiě)出拋物線的標(biāo)準(zhǔn)方程：（1）焦點(diǎn)是SKIPIF1<0；（2）準(zhǔn)線方程是SKIPIF1<0；（3）焦點(diǎn)到準(zhǔn)線的距離是SKIPIF1<0．【答案】（1）SKIPIF1<0；（2）SKIPIF1<0；（3）SKIPIF1<0或SKIPIF1<0.【詳解】（1）由題意可知拋物線的焦點(diǎn)在SKIPIF1<0軸的正半軸上，設(shè)拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，所以，拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0；（2）由題意可知拋物線的焦點(diǎn)在SKIPIF1<0軸的正半軸上，設(shè)拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，因此，拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0；（3）拋物線的焦點(diǎn)到準(zhǔn)線的距離為SKIPIF1<0，所以，拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0或SKIPIF1<0.【變式1】（2023秋·陜西西安·高二?？计谀?duì)拋物線SKIPIF1<0，下列描述正確的是A．開(kāi)口向上，焦點(diǎn)為SKIPIF1<0 B．開(kāi)口向上，焦點(diǎn)為SKIPIF1<0C．開(kāi)口向右，焦點(diǎn)為SKIPIF1<0 D．開(kāi)口向右，焦點(diǎn)為SKIPIF1<0【答案】B【詳解】解：因?yàn)閽佄锞€SKIPIF1<0，可知化為標(biāo)準(zhǔn)式為拋物線SKIPIF1<0，2p=1/4，故焦點(diǎn)在y軸上，開(kāi)口向上，焦點(diǎn)坐標(biāo)為SKIPIF1<0，選B【變式2】（2023春·湖南長(zhǎng)沙·高二長(zhǎng)沙市明德中學(xué)校考期中）若拋物線SKIPIF1<0的焦點(diǎn)與雙曲線SKIPIF1<0的右焦點(diǎn)重合，則SKIPIF1<0的值．【答案】6【詳解】試題分析：根據(jù)題意，由于雙曲線SKIPIF1<0的SKIPIF1<0右焦點(diǎn)坐標(biāo)為SKIPIF1<0，因此可知拋物線SKIPIF1<0的焦點(diǎn)SKIPIF1<0,故答案為6題型02直線與拋物線的位置關(guān)系【典例1】（2023秋·高二課時(shí)練習(xí)）已知直線SKIPIF1<0，拋物線SKIPIF1<0，l與SKIPIF1<0有一個(gè)公共點(diǎn)的直線有（

）A．1條 B．2條 C．3條D．1條、2條或3條【答案】C【詳解】聯(lián)立直線SKIPIF1<0和拋物線SKIPIF1<0方程可得SKIPIF1<0，整理可得SKIPIF1<0，直線l與SKIPIF1<0有一個(gè)公共點(diǎn)等價(jià)于方程只有一個(gè)實(shí)數(shù)根，當(dāng)SKIPIF1<0時(shí)，方程為SKIPIF1<0僅有一解，符合題意；當(dāng)SKIPIF1<0時(shí)，一元二次方程SKIPIF1<0僅有一解，即SKIPIF1<0，解得SKIPIF1<0，所以滿足題意得直線有三條，即SKIPIF1<0，SKIPIF1<0和SKIPIF1<0.故選：C【典例2】（多選）（2023·全國(guó)·高三專(zhuān)題練習(xí)）若經(jīng)過(guò)點(diǎn)SKIPIF1<0的直線與拋物線SKIPIF1<0恒有公共點(diǎn)，則C的準(zhǔn)線可能是（

）．A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BD【詳解】由題意得，點(diǎn)SKIPIF1<0在拋物線上或其內(nèi)部，則SKIPIF1<0，解得SKIPIF1<0，∴其準(zhǔn)線為SKIPIF1<0.故選:BD．【典例3】（2023春·湖北孝感·高二校聯(lián)考階段練習(xí)）已知M是拋物線SKIPIF1<0上一點(diǎn)，則點(diǎn)M到直線SKIPIF1<0的最短距離為．【答案】SKIPIF1<0/SKIPIF1<0【詳解】設(shè)SKIPIF1<0，則點(diǎn)M到直線SKIPIF1<0的距離SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)取等號(hào).故答案為：SKIPIF1<0【典例4】（2023秋·廣西北?！じ叨y(tǒng)考期末）已知拋物線SKIPIF1<0，其準(zhǔn)線方程為SKIPIF1<0．(1)求拋物線SKIPIF1<0的方程；(2)不過(guò)原點(diǎn)SKIPIF1<0的直線SKIPIF1<0與拋物線交于不同的兩點(diǎn)SKIPIF1<0，且SKIPIF1<0，求SKIPIF1<0的值．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）準(zhǔn)線為SKIPIF1<0，SKIPIF1<0，拋物線SKIPIF1<0的方程為SKIPIF1<0；（2）設(shè)SKIPIF1<0，聯(lián)立SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，經(jīng)檢驗(yàn)，當(dāng)SKIPIF1<0時(shí)，直線過(guò)坐標(biāo)原點(diǎn)，不合題意，又SKIPIF1<0，符合題意；綜上，m的值為SKIPIF1<0．【變式1】（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知拋物線SKIPIF1<0與直線SKIPIF1<0有且僅有一個(gè)交點(diǎn)，則SKIPIF1<0（

）A．4 B．2 C．0或4 D．8【答案】C【詳解】聯(lián)立SKIPIF1<0得：SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，交點(diǎn)為SKIPIF1<0，滿足題意；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，解得SKIPIF1<0，綜上可知：SKIPIF1<0或SKIPIF1<0，故選：C【變式2】（多選）（2023秋·安徽阜陽(yáng)·高二統(tǒng)考期末）若直線SKIPIF1<0與拋物線SKIPIF1<0只有一個(gè)交點(diǎn)，則SKIPIF1<0的可能取值為（

）A．2 B．SKIPIF1<0 C．SKIPIF1<0 D．0【答案】BD【詳解】聯(lián)立SKIPIF1<0，消去SKIPIF1<0可得SKIPIF1<0，∵直線SKIPIF1<0與拋物線SKIPIF1<0只有一個(gè)交點(diǎn)，SKIPIF1<0或SKIPIF1<0.故選：BD.【變式3】（2023秋·廣東廣州·高二?？计谀┮阎獟佄锞€SKIPIF1<0的一條切線方程為SKIPIF1<0，則SKIPIF1<0的準(zhǔn)線方程為．【答案】SKIPIF1<0【詳解】由SKIPIF1<0，消去SKIPIF1<0得SKIPIF1<0，由題意SKIPIF1<0，解得SKIPIF1<0，則拋物線方程為：SKIPIF1<0，所以拋物線的準(zhǔn)線方程為：SKIPIF1<0，即SKIPIF1<0．故答案為：SKIPIF1<0．【變式4】（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知橢圓SKIPIF1<0，設(shè)直線l同時(shí)與橢圓和拋物線SKIPIF1<0各恰有一個(gè)公共交點(diǎn)，求直線l的方程．【答案】SKIPIF1<0或SKIPIF1<0【詳解】由題，直線SKIPIF1<0的斜率存在，并設(shè)方程為SKIPIF1<0，聯(lián)立SKIPIF1<0整理得SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，整理得SKIPIF1<0，聯(lián)立SKIPIF1<0整理得SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0，則有SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0解得SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，所以直線SKIPIF1<0的方程為SKIPIF1<0或SKIPIF1<0.題型03拋物線的弦長(zhǎng)【典例1】（2023秋·浙江寧波·高二統(tǒng)考期末）已知拋物線SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0的直線SKIPIF1<0交拋物線于SKIPIF1<0兩點(diǎn)，且弦SKIPIF1<0被點(diǎn)SKIPIF1<0平分.(1)求直線SKIPIF1<0的方程；(2)求弦SKIPIF1<0的長(zhǎng)度.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）設(shè)SKIPIF1<0則SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0所以SKIPIF1<0，得直線SKIPIF1<0的方程為SKIPIF1<0.（2）聯(lián)立方程SKIPIF1<0，得SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0【典例2】（2023秋·高二課時(shí)練習(xí)）直線SKIPIF1<0與拋物線SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，求線段AB的長(zhǎng)．【答案】SKIPIF1<0．【詳解】解：拋物線SKIPIF1<0，直線SKIPIF1<0，將直線方程代入到拋物線方程中，得：SKIPIF1<0，整理得：SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，由一元二次方程根與系數(shù)的關(guān)系得：SKIPIF1<0，SKIPIF1<0，所以弦長(zhǎng)SKIPIF1<0．【變式1】（2023春·安徽滁州·高二?？奸_(kāi)學(xué)考試）已知?jiǎng)訄ASKIPIF1<0過(guò)定點(diǎn)SKIPIF1<0，且與直線SKIPIF1<0：SKIPIF1<0相切，圓心SKIPIF1<0的軌跡為SKIPIF1<0．(1)求動(dòng)點(diǎn)SKIPIF1<0的軌跡方程；(2)過(guò)點(diǎn)SKIPIF1<0作傾斜角為SKIPIF1<0的直線SKIPIF1<0交軌跡SKIPIF1<0于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，求SKIPIF1<0．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）設(shè)SKIPIF1<0，由動(dòng)圓SKIPIF1<0過(guò)定點(diǎn)SKIPIF1<0，且與直線SKIPIF1<0：SKIPIF1<0相切，SKIPIF1<0，整理得SKIPIF1<0，故動(dòng)點(diǎn)SKIPIF1<0的軌跡方程為SKIPIF1<0．（2）設(shè)SKIPIF1<0，SKIPIF1<0，直線SKIPIF1<0的方程為SKIPIF1<0，則由SKIPIF1<0，整理得SKIPIF1<0,SKIPIF1<0SKIPIF1<0．【變式2】（2023春·四川成都·高二成都外國(guó)語(yǔ)學(xué)校校考階段練習(xí)）已知拋物線SKIPIF1<0的準(zhǔn)線方程為SKIPIF1<0.(1)求SKIPIF1<0的值；(2)直線SKIPIF1<0交拋物線于SKIPIF1<0、SKIPIF1<0兩點(diǎn)，求弦長(zhǎng)SKIPIF1<0.【答案】(1)2；(2)SKIPIF1<0.【詳解】（1）拋物線SKIPIF1<0的準(zhǔn)線方程為SKIPIF1<0，依題意，SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的值為2.（2）由（1）知，拋物線SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0消去y得：SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.題型04拋物線的中點(diǎn)弦和點(diǎn)差法【典例1】（2023秋·陜西咸陽(yáng)·高二?？计谀┮阎獟佄锞€SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0引拋物線的一條弦，使它恰在點(diǎn)SKIPIF1<0處被平分，則這條弦所在的直線SKIPIF1<0的方程為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】易知直線l的斜率存在，設(shè)直線的斜率為k，直線l交拋物線于M，N兩點(diǎn)，設(shè)SKIPIF1<0，則SKIPIF1<0，兩式相減得SKIPIF1<0，整理得SKIPIF1<0，因?yàn)镸N的中點(diǎn)為SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，所以直線l的方程為SKIPIF1<0即SKIPIF1<0.故選：A【典例2】（2023春·寧夏吳忠·高二吳忠中學(xué)校考期中）已知拋物線SKIPIF1<0是拋物線SKIPIF1<0上的點(diǎn)，且SKIPIF1<0.(1)求拋物線SKIPIF1<0的方程；(2)已知直線SKIPIF1<0交拋物線SKIPIF1<0于SKIPIF1<0兩點(diǎn)，且SKIPIF1<0的中點(diǎn)為SKIPIF1<0，求直線SKIPIF1<0的方程.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）由題意，在拋物線SKIPIF1<0中，SKIPIF1<0，由幾何知識(shí)得，SKIPIF1<0，解得：SKIPIF1<0，故拋物線SKIPIF1<0的方程為：SKIPIF1<0.（2）由題意及（1）得，直線SKIPIF1<0的斜率存在，設(shè)直線SKIPIF1<0的斜率為SKIPIF1<0，則SKIPIF1<0，兩式相減得SKIPIF1<0，整理得SKIPIF1<0，因?yàn)镾KIPIF1<0的中點(diǎn)為SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0，∴直線SKIPIF1<0的方程為：SKIPIF1<0，即SKIPIF1<0，經(jīng)檢驗(yàn)，滿足題意.【變式1】（2023秋·甘肅慶陽(yáng)·高二校考期末）已知點(diǎn)SKIPIF1<0，若拋物線SKIPIF1<0的一條弦AB恰好是以P為中點(diǎn)，則弦AB所在直線方程是．【答案】SKIPIF1<0【詳解】SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，SKIPIF1<0在拋物線內(nèi)部（含焦點(diǎn)的部分），設(shè)SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0，相減得SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，直線SKIPIF1<0方程為SKIPIF1<0，即SKIPIF1<0，故答案為：SKIPIF1<0．【變式2】（2023·江蘇·高二專(zhuān)題練習(xí)）已知頂點(diǎn)在原點(diǎn)，焦點(diǎn)在SKIPIF1<0軸上的拋物線過(guò)點(diǎn)SKIPIF1<0．(1)求拋物線的標(biāo)準(zhǔn)方程；(2)過(guò)點(diǎn)SKIPIF1<0作直線交拋物線于A、B兩點(diǎn)，使得Q恰好平分線段AB，求直線AB的方程．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）因?yàn)轫旤c(diǎn)在原點(diǎn)，焦點(diǎn)在y軸上的拋物線過(guò)點(diǎn)SKIPIF1<0，所以拋物線的焦點(diǎn)在y軸正半軸，設(shè)其方程為SKIPIF1<0，將點(diǎn)SKIPIF1<0代入可得SKIPIF1<0，所以SKIPIF1<0，所以拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0，（2）拋物線SKIPIF1<0中，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在拋物線內(nèi)部，可以為弦的中點(diǎn)．設(shè)點(diǎn)SKIPIF1<0，直線SKIPIF1<0斜率為SKIPIF1<0點(diǎn)SKIPIF1<0在拋物線上，所以SKIPIF1<0所以SKIPIF1<0，即SKIPIF1<0，所以直線方程為SKIPIF1<0．經(jīng)檢驗(yàn)，直線SKIPIF1<0符合題意.題型05拋物線的焦點(diǎn)弦【典例1】（2023·遼寧朝陽(yáng)·朝陽(yáng)市第一高級(jí)中學(xué)?？寄M預(yù)測(cè)）過(guò)拋物線SKIPIF1<0：SKIPIF1<0焦點(diǎn)SKIPIF1<0的直線與SKIPIF1<0交于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，過(guò)點(diǎn)SKIPIF1<0向拋物線SKIPIF1<0的準(zhǔn)線作垂線，垂足為SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．18 D．20【答案】B【詳解】依題意拋物線的準(zhǔn)線為SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以拋物線方程為SKIPIF1<0，則焦點(diǎn)為SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以直線SKIPIF1<0的方程為SKIPIF1<0，由SKIPIF1<0，消去SKIPIF1<0整理得SKIPIF1<0，解得SKIPIF1<0、SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.故選：B【典例2】（2023春·湖北孝感·高二統(tǒng)考開(kāi)學(xué)考試）已知曲線C位于y軸右側(cè)，且曲線C上任意一點(diǎn)P與定點(diǎn)SKIPIF1<0的距離比它到y(tǒng)軸的距離大1．(1)求曲線C的軌跡方程；(2)若直線l經(jīng)過(guò)點(diǎn)F，與曲線C交于A，B兩點(diǎn)，且SKIPIF1<0，求直線l的方程．【答案】(1)SKIPIF1<0；(2)SKIPIF1<0或SKIPIF1<0．【詳解】（1）由題意動(dòng)點(diǎn)SKIPIF1<0與定點(diǎn)SKIPIF1<0的距離和它到直線SKIPIF1<0的距離相等，所以，曲線C是以F為焦點(diǎn)，直線SKIPIF1<0為準(zhǔn)線的拋物線（去掉頂點(diǎn)），SKIPIF1<0，所以曲線C的軌跡方程是SKIPIF1<0；（2）若直線SKIPIF1<0斜率不存在，則SKIPIF1<0不合題意，因此直線SKIPIF1<0斜率存在，設(shè)直線SKIPIF1<0方程為SKIPIF1<0，代入曲線C方程整理得SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，所以直線SKIPIF1<0方程為SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0．【典例3】（2023·全國(guó)·模擬預(yù)測(cè)）已知點(diǎn)SKIPIF1<0在拋物線SKIPIF1<0上，記SKIPIF1<0為坐標(biāo)原點(diǎn)，SKIPIF1<0，以SKIPIF1<0為圓心，SKIPIF1<0為半徑的圓與拋物線SKIPIF1<0的準(zhǔn)線相切．(1)求拋物線SKIPIF1<0的方程；(2)記拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作直線SKIPIF1<0與直線SKIPIF1<0垂直，交拋物線SKIPIF1<0于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，求弦SKIPIF1<0的長(zhǎng)．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，準(zhǔn)線方程為SKIPIF1<0，依題意可得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，又SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，所以SKIPIF1<0，所以拋物線方程為SKIPIF1<0.（2）由（1）可得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，因?yàn)橹本€SKIPIF1<0直線SKIPIF1<0，所以SKIPIF1<0，所以直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0，由SKIPIF1<0，消去SKIPIF1<0整理得SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.【變式1】（2023春·上海寶山·高三上海交大附中?？计谥校┻^(guò)拋物線SKIPIF1<0的焦點(diǎn)且傾斜角為SKIPIF1<0的直線被拋物線截得的弦長(zhǎng)為.【答案】SKIPIF1<0【詳解】拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，準(zhǔn)線方程為SKIPIF1<0，直線SKIPIF1<0的傾斜角為SKIPIF1<0，設(shè)直線SKIPIF1<0與拋物線交于SKIPIF1<0兩點(diǎn)，則直線SKIPIF1<0的方程為SKIPIF1<0，代入SKIPIF1<0得SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故答案為：SKIPIF1<0【變式2】（2023春·廣東汕尾·高二統(tǒng)考期末）已知拋物線SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0（SKIPIF1<0）．(1)求C的方程；(2)若斜率為SKIPIF1<0的直線過(guò)C的焦點(diǎn)，且與C交于A，B兩點(diǎn)，求線段SKIPIF1<0的長(zhǎng)度．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）∵拋物線SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0，∴SKIPIF1<0．又∵SKIPIF1<0，∴SKIPIF1<0，上故SKIPIF1<0的方程為SKIPIF1<0．（2）設(shè)SKIPIF1<0，SKIPIF1<0，由（1）知，拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，∵直線SKIPIF1<0的斜率為SKIPIF1<0，且過(guò)點(diǎn)SKIPIF1<0，∴直線SKIPIF1<0的方程為SKIPIF1<0，

聯(lián)立SKIPIF1<0得SKIPIF1<0，則SKIPIF1<0．

∴SKIPIF1<0，故線段SKIPIF1<0的長(zhǎng)度為SKIPIF1<0．【變式3】（2023春·貴州黔東南·高二?？茧A段練習(xí)）已知拋物線SKIPIF1<0的焦點(diǎn)SKIPIF1<0關(guān)于拋物線SKIPIF1<0的準(zhǔn)線的對(duì)稱(chēng)點(diǎn)為SKIPIF1<0．(1)求拋物線SKIPIF1<0的方程；(2)過(guò)點(diǎn)SKIPIF1<0作斜率為4直線SKIPIF1<0，交拋物線SKIPIF1<0于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，求SKIPIF1<0．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）該拋物線的焦點(diǎn)坐標(biāo)為SKIPIF1<0，準(zhǔn)線方程為SKIPIF1<0，因?yàn)镾KIPIF1<0關(guān)于拋物線SKIPIF1<0的準(zhǔn)線的對(duì)稱(chēng)點(diǎn)為SKIPIF1<0，所以有SKIPIF1<0；（2）直線SKIPIF1<0的方程為SKIPIF1<0，與拋物線方程聯(lián)立，得SKIPIF1<0，設(shè)SKIPIF1<0，因此有SKIPIF1<0，則有SKIPIF1<0題型06拋物線的定值、定點(diǎn)、定直線問(wèn)題【典例1】（2023春·四川資陽(yáng)·高二統(tǒng)考期末）過(guò)點(diǎn)SKIPIF1<0作拋物線SKIPIF1<0在第一象限部分的切線，切點(diǎn)為A，F(xiàn)為SKIPIF1<0的焦點(diǎn)，SKIPIF1<0為坐標(biāo)原點(diǎn)，SKIPIF1<0的面積為1．(1)求SKIPIF1<0的方程；(2)過(guò)點(diǎn)SKIPIF1<0作兩條互相垂直的直線SKIPIF1<0和SKIPIF1<0，SKIPIF1<0交SKIPIF1<0于C，D兩點(diǎn)，SKIPIF1<0交SKIPIF1<0于P，Q兩點(diǎn)，且M，N分別為線段CD和PQ的中點(diǎn)．直線MN是否恒過(guò)一個(gè)定點(diǎn)？若是，求出該定點(diǎn)坐標(biāo)；若不是，說(shuō)明理由．【答案】(1)SKIPIF1<0(2)直線MN恒過(guò)定點(diǎn)SKIPIF1<0．【詳解】（1）由題，SKIPIF1<0，設(shè)切點(diǎn)SKIPIF1<0，則切線方程為SKIPIF1<0，SKIPIF1<0,SKIPIF1<0的坐標(biāo)代入，得SKIPIF1<0，解得SKIPIF1<0，由于SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0的面積SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的方程為SKIPIF1<0．（2）由題意可知，直線SKIPIF1<0和SKIPIF1<0斜率都存在且均不為0，設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，則直線SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立方程組SKIPIF1<0消去SKIPIF1<0并整理得，SKIPIF1<0，則SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0為CD中點(diǎn)，所以SKIPIF1<0，同理可得SKIPIF1<0，所以，直線MN的方程為SKIPIF1<0，整理得SKIPIF1<0，所以，直線MN恒過(guò)定點(diǎn)SKIPIF1<0．【典例2】（2023·河南信陽(yáng)·信陽(yáng)高中?？既＃┮阎獟佄锞€SKIPIF1<0上一點(diǎn)SKIPIF1<0到焦點(diǎn)的距離為3.

(1)求SKIPIF1<0，SKIPIF1<0的值；(2)設(shè)SKIPIF1<0為直線SKIPIF1<0上除SKIPIF1<0，SKIPIF1<0兩點(diǎn)外的任意一點(diǎn)，過(guò)SKIPIF1<0作圓SKIPIF1<0的兩條切線，分別與曲線SKIPIF1<0相交于點(diǎn)SKIPIF1<0，SKIPIF1<0和SKIPIF1<0，SKIPIF1<0，試判斷SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0四點(diǎn)縱坐標(biāo)之積是否為定值？若是，求該定值；若不是，請(qǐng)說(shuō)明理由.【答案】(1)SKIPIF1<0，SKIPIF1<0(2)定值為64【詳解】（1）根據(jù)拋物線的定義，SKIPIF1<0到準(zhǔn)線SKIPIF1<0的距離為3，∴SKIPIF1<0，∴SKIPIF1<0；∴拋物線的焦點(diǎn)坐標(biāo)為SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0；（2）設(shè)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0的直線方程設(shè)為SKIPIF1<0，由SKIPIF1<0得，SKIPIF1<0，若直線SKIPIF1<0，SKIPIF1<0的斜率分別為SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的縱坐標(biāo)分別為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0到SKIPIF1<0的距離SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0四點(diǎn)縱坐標(biāo)之積為定值，且定值為64.【典例3】（2023·廣西·統(tǒng)考一模）已知拋物線SKIPIF1<0和圓SKIPIF1<0，傾斜角為45°的直線SKIPIF1<0過(guò)SKIPIF1<0的焦點(diǎn)且與SKIPIF1<0相切．(1)求p的值：(2)點(diǎn)M在SKIPIF1<0的準(zhǔn)線上，動(dòng)點(diǎn)A在SKIPIF1<0上，SKIPIF1<0在A點(diǎn)處的切線l2交y軸于點(diǎn)B，設(shè)SKIPIF1<0，求證：點(diǎn)N在定直線上，并求該定直線的方程．【答案】(1)SKIPIF1<0；(2)證明見(jiàn)解析，定直線方程為SKIPIF1<0.【詳解】（1）由題得拋物線SKIPIF1<0的焦點(diǎn)坐標(biāo)為SKIPIF1<0，設(shè)直線l1的方程為SKIPIF1<0，由已知得圓SKIPIF1<0的圓心SKIPIF1<0，半徑SKIPIF1<0，因?yàn)橹本€l1與圓SKIPIF1<0相切，所以圓心到直線SKIPIF1<0的距離SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去）．所以SKIPIF1<0.（2）依題意設(shè)SKIPIF1<0，由（1）知拋物線SKIPIF1<0方程為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，設(shè)ASKIPIF1<0，SKIPIF1<0），則以A為切點(diǎn)的切線l2的斜率為SKIPIF1<0所以切線l2的方程為SKIPIF1<0．令SKIPIF1<0，即l2交y軸于B點(diǎn)坐標(biāo)為SKIPIF1<0，所以SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0．設(shè)N點(diǎn)坐標(biāo)為（x，y），則SKIPIF1<0，所以點(diǎn)N在定直線SKIPIF1<0上．

【變式1】（2023春·河北·高二校聯(lián)考期末）已知SKIPIF1<0為拋物線SKIPIF1<0上一點(diǎn)，SKIPIF1<0，SKIPIF1<0為SKIPIF1<0的中點(diǎn)，設(shè)SKIPIF1<0的軌跡為曲線SKIPIF1<0．(1)求曲線SKIPIF1<0的方程；(2)過(guò)點(diǎn)SKIPIF1<0作直線交曲線E于點(diǎn)M、N，點(diǎn)SKIPIF1<0為直線l：SKIPIF1<0上一動(dòng)點(diǎn)．問(wèn)是否存在點(diǎn)SKIPIF1<0使SKIPIF1<0為正三角形？若存在，求出點(diǎn)SKIPIF1<0坐標(biāo)；若不存在，請(qǐng)說(shuō)明理由．【答案】(1)SKIPIF1<0(2)存在；SKIPIF1<0【詳解】（1）設(shè)SKIPIF1<0，則SKIPIF1<0因?yàn)辄c(diǎn)B在拋物線SKIPIF1<0上，即SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0，所以曲線E的方程為SKIPIF1<0．（2）假設(shè)存在點(diǎn)SKIPIF1<0使SKIPIF1<0為正三角形．當(dāng)MN垂直于y軸時(shí)，不符合題意；當(dāng)MN不垂直于y軸時(shí)，設(shè)直線MN：SKIPIF1<0，MN的中點(diǎn)為SKIPIF1<0，聯(lián)立SKIPIF1<0得：SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0為正三角形，∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，PK：SKIPIF1<0，令SKIPIF1<0，∴SKIPIF1<0所以存在點(diǎn)SKIPIF1<0使SKIPIF1<0為正三角形．

【變式2】（2023·陜西咸陽(yáng)·武功縣普集高級(jí)中學(xué)?？寄M預(yù)測(cè)）已知點(diǎn)SKIPIF1<0是拋物線SKIPIF1<0：SKIPIF1<0的焦點(diǎn)，縱坐標(biāo)為2的點(diǎn)SKIPIF1<0在SKIPIF1<0上，以SKIPIF1<0為圓心、SKIPIF1<0為半徑的圓交SKIPIF1<0軸于SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.(1)求拋物線SKIPIF1<0的方程；(2)過(guò)SKIPIF1<0作直線SKIPIF1<0與拋物線SKIPIF1<0交于SKIPIF1<0，SKIPIF1<0，求SKIPIF1<0的值.【答案】(1)SKIPIF1<0(2)2【詳解】（1）由題知，SKIPIF1<0點(diǎn)的橫坐標(biāo)為SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，解得SKIPIF1<0，∴拋物線SKIPIF1<0的方程為SKIPIF1<0.

（2）由（1）知SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，直線SKIPIF1<0的方程為SKIPIF1<0，代入SKIPIF1<0，整理得SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0SKIPIF1<0SKIPIF1<0.

【變式3】（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知拋物線E：SKIPIF1<0（p>0），過(guò)點(diǎn)SKIPIF1<0的兩條直線l1，l2分別交E于AB兩點(diǎn)和C，D兩點(diǎn)．當(dāng)l1的斜率為SKIPIF1<0時(shí)，SKIPIF1<0(1)求E的標(biāo)準(zhǔn)方程：(2)設(shè)G為直線AD與BC的交點(diǎn)，證明：點(diǎn)G必在定直線上．【答案】(1)SKIPIF1<0(2)證明見(jiàn)解析【詳解】（1）當(dāng)SKIPIF1<0的斜率為SKIPIF1<0時(shí)，得SKIPIF1<0方程為SKIPIF1<0，由SKIPIF1<0,消元得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；由弦長(zhǎng)公式得SKIPIF1<0SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去），SKIPIF1<0滿足SKIPIF1<0，從而SKIPIF1<0的標(biāo)準(zhǔn)方程為SKIPIF1<0.（2）法一：因?yàn)閘1，l2分別交E于AB兩點(diǎn)和C，D兩點(diǎn)，所以直線斜率存在設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，設(shè)SKIPIF1<0，由SKIPIF1<0，消去SKIPIF1<0得SKIPIF1<0，則SKIPIF1<0.設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，同理SKIPIF1<0，消去SKIPIF1<0得SKIPIF1<0可得SKIPIF1<0．直線SKIPIF1<0方程為SKIPIF1<0，即SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0，同理，直線SKIPIF1<0方程為