人教A版高中數(shù)學(xué)(選擇性必修第一冊(cè))同步講義第30講 3.3.1拋物線及其標(biāo)準(zhǔn)方程（含解析）

第05講3.3.1拋物線及其標(biāo)準(zhǔn)方程課程標(biāo)準(zhǔn)學(xué)習(xí)目標(biāo)①掌握拋物線的定義、標(biāo)準(zhǔn)方程和拋物線的簡(jiǎn)單性質(zhì)。②了解拋物線在實(shí)際問題中的初步應(yīng)用。通過本節(jié)課的學(xué)習(xí)，要求掌握拋物線的定義，標(biāo)準(zhǔn)方程及相關(guān)的條件，并能應(yīng)用拋物線的定義解決實(shí)際問題知識(shí)點(diǎn)01：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).【即學(xué)即練1】（2023春·四川涼山·高二寧南中學(xué)校聯(lián)考期末）已知拋物線SKIPIF1<0上一點(diǎn)P到y(tǒng)軸的距離為2，焦點(diǎn)為F，則SKIPIF1<0（

）A．2 B．3 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】由題得拋物線的準(zhǔn)線方程為SKIPIF1<0，所以點(diǎn)P到準(zhǔn)線的距離為SKIPIF1<0，由拋物線的定義得SKIPIF1<03.故選：B

知識(shí)點(diǎn)02：拋物線的標(biāo)準(zhǔn)方程設(shè)SKIPIF1<0，拋物線的標(biāo)準(zhǔn)方程、類型及其幾何性質(zhì)：方程SKIPIF1<0（SKIPIF1<0）SKIPIF1<0（SKIPIF1<0）SKIPIF1<0（SKIPIF1<0）SKIPIF1<0（SKIPIF1<0）圖形焦點(diǎn)SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0準(zhǔn)線SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0【即學(xué)即練2】（2023秋·高二課時(shí)練習(xí)）已知拋物線的標(biāo)準(zhǔn)方程如下，分別求其焦點(diǎn)和準(zhǔn)線方程：(1)SKIPIF1<0；(2)SKIPIF1<0．【答案】(1)焦點(diǎn)為SKIPIF1<0，準(zhǔn)線方程為SKIPIF1<0；(2)焦點(diǎn)為SKIPIF1<0，準(zhǔn)線方程為SKIPIF1<0．【詳解】（1）由拋物線方程為SKIPIF1<0，可得SKIPIF1<0，且焦點(diǎn)在SKIPIF1<0軸正半軸上，所以可得其焦點(diǎn)為SKIPIF1<0，準(zhǔn)線方程為SKIPIF1<0；（2）將SKIPIF1<0化成標(biāo)準(zhǔn)方程為SKIPIF1<0，可得SKIPIF1<0，且焦點(diǎn)在SKIPIF1<0軸負(fù)半軸上，所以焦點(diǎn)為SKIPIF1<0，準(zhǔn)線方程為SKIPIF1<0.特別說明：1、要注意弄清拋物線四種形式的標(biāo)準(zhǔn)方程的特征及其對(duì)應(yīng)拋物線的形狀(焦點(diǎn)位置、開口方向等).拋物線的標(biāo)準(zhǔn)方程中,有一個(gè)一次項(xiàng)和一個(gè)二次項(xiàng),二次項(xiàng)的系數(shù)為1,一次項(xiàng)的系數(shù)為SKIPIF1<0;若一次項(xiàng)的字母是SKIPIF1<0,則焦點(diǎn)就在SKIPIF1<0軸上,若其系數(shù)是正的,則焦點(diǎn)就在SKIPIF1<0軸的正半軸上(開口向右),若系數(shù)是負(fù)的,焦點(diǎn)就在SKIPIF1<0軸的負(fù)半軸上(開口向左);若一次項(xiàng)的字母是SKIPIF1<0,則焦點(diǎn)就在SKIPIF1<0軸上,若其系數(shù)是正的,則焦點(diǎn)就在SKIPIF1<0軸的正半軸上(開口向上),若系數(shù)是負(fù)的,焦點(diǎn)就在SKIPIF1<0軸的負(fù)半軸上(開口向下).2、焦點(diǎn)的非零坐標(biāo)是標(biāo)準(zhǔn)方程下一次項(xiàng)系數(shù)的.3、準(zhǔn)線與坐標(biāo)軸的交點(diǎn)和拋物線的焦點(diǎn)關(guān)于原點(diǎn)對(duì)稱.4、（1）通徑：過焦點(diǎn)且垂直于對(duì)稱軸的弦長(zhǎng)等于SKIPIF1<0，通徑是過焦點(diǎn)最短的弦.（2）拋物線SKIPIF1<0（SKIPIF1<0）上一點(diǎn)SKIPIF1<0到焦點(diǎn)SKIPIF1<0的距離SKIPIF1<0，也稱為拋物線的焦半徑.題型01拋物線定義的理解【典例1】（2023秋·陜西西安·高二統(tǒng)考期末）若拋物線SKIPIF1<0上一點(diǎn)SKIPIF1<0到SKIPIF1<0軸的距離為SKIPIF1<0，則點(diǎn)SKIPIF1<0到拋物線的焦點(diǎn)SKIPIF1<0的距離為．【答案】4【詳解】由題意可得，SKIPIF1<0，P縱坐標(biāo)為SKIPIF1<0，由其解析式可得P橫坐標(biāo)為SKIPIF1<0，由拋物線定義知SKIPIF1<0.故答案為：4【典例2】（2023·四川成都·四川省成都列五中學(xué)校考三模）若拋物線SKIPIF1<0上的點(diǎn)P到焦點(diǎn)的距離為8，到SKIPIF1<0軸的距離為6，則拋物線SKIPIF1<0的標(biāo)準(zhǔn)方程是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】由拋物線定義可得：SKIPIF1<0，解得SKIPIF1<0，所以拋物線SKIPIF1<0的標(biāo)準(zhǔn)方程為SKIPIF1<0.故選：C

【變式1】（2023春·陜西榆林·高二統(tǒng)考期末）已知拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，點(diǎn)SKIPIF1<0在SKIPIF1<0上，若SKIPIF1<0到直線SKIPIF1<0的距離為7，則SKIPIF1<0.【答案】SKIPIF1<0【詳解】由拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，準(zhǔn)線方程為SKIPIF1<0，因?yàn)辄c(diǎn)SKIPIF1<0在SKIPIF1<0上，且SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0，可得SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0，即點(diǎn)SKIPIF1<0到準(zhǔn)線的距離為SKIPIF1<0，根據(jù)拋物線的定義，可得點(diǎn)SKIPIF1<0到焦點(diǎn)的距離等于點(diǎn)SKIPIF1<0到準(zhǔn)線的距離，所以SKIPIF1<0.故答案為：SKIPIF1<0.【變式2】（2023春·江西宜春·高三江西省宜春中學(xué)?？茧A段練習(xí)）若拋物線SKIPIF1<0上一點(diǎn)SKIPIF1<0到焦點(diǎn)的距離是該點(diǎn)到SKIPIF1<0軸距離的3倍，則SKIPIF1<0.【答案】SKIPIF1<0/3.5【詳解】由題知：SKIPIF1<0，故由焦半徑公式得：SKIPIF1<0.故答案為：SKIPIF1<0.題型02利用拋物線定義求方程【典例1】（2023春·江西·高三校聯(lián)考階段練習(xí)）設(shè)圓SKIPIF1<0與y軸交于A，B兩點(diǎn)（A在B的上方），過B作圓O的切線l，若動(dòng)點(diǎn)P到A的距離等于P到l的距離，則動(dòng)點(diǎn)P的軌跡方程為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因?yàn)閳ASKIPIF1<0與SKIPIF1<0軸交于SKIPIF1<0，SKIPIF1<0兩點(diǎn)（SKIPIF1<0在SKIPIF1<0的上方），所以SKIPIF1<0，SKIPIF1<0，又因?yàn)檫^SKIPIF1<0作圓SKIPIF1<0的切線SKIPIF1<0，所以切線SKIPIF1<0的方程為SKIPIF1<0，因?yàn)閯?dòng)點(diǎn)SKIPIF1<0到SKIPIF1<0的距離等于SKIPIF1<0到SKIPIF1<0的距離，所以動(dòng)點(diǎn)SKIPIF1<0的軌跡為拋物線，且其焦點(diǎn)為SKIPIF1<0，準(zhǔn)線為SKIPIF1<0，所以SKIPIF1<0的軌跡方程為SKIPIF1<0．故選：A.【典例2】（2023·全國(guó)·高三專題練習(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．【答案】SKIPIF1<0【詳解】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【變式1】（2023·全國(guó)·高三專題練習(xí)）已知點(diǎn)SKIPIF1<0，過點(diǎn)SKIPIF1<0且與y軸垂直的直線為SKIPIF1<0，SKIPIF1<0軸，交SKIPIF1<0于點(diǎn)N，直線l垂直平分FN，交SKIPIF1<0于點(diǎn)M.求點(diǎn)M的軌跡方程；【答案】SKIPIF1<0【詳解】由題意得SKIPIF1<0，即動(dòng)點(diǎn)M到點(diǎn)SKIPIF1<0的距離和到直線SKIPIF1<0的距離相等，所以點(diǎn)M的軌跡是以SKIPIF1<0為焦點(diǎn)，直線SKIPIF1<0為準(zhǔn)線的拋物線，根據(jù)拋物線定義可知點(diǎn)M的軌跡方程為SKIPIF1<0；【變式2】（2023·全國(guó)·高三專題練習(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．【詳解】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．題型03拋物線上點(diǎn)到定點(diǎn)距離及最值【典例1】（2023春·河南焦作·高二統(tǒng)考開學(xué)考試）已知點(diǎn)A是拋物線SKIPIF1<0上的點(diǎn)，點(diǎn)SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．2 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最小值SKIPIF1<0．故選：A【典例2】（2023春·云南昭通·高三校考階段練習(xí)）拋物線SKIPIF1<0上任意一點(diǎn)P到點(diǎn)SKIPIF1<0的距離最小值為.【答案】SKIPIF1<0【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)取得最小值4，故答案為：4【變式1】（2023·全國(guó)·高三專題練習(xí)）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【詳解】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【變式2】（2023·全國(guó)·高三專題練習(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.【答案】3【詳解】因?yàn)閽佄锞€和圓都關(guān)于橫軸對(duì)稱，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<0題型04拋物線上點(diǎn)到定點(diǎn)與焦點(diǎn)距離的和（差）最值【典例1】（2023秋·陜西·高二校聯(lián)考期末）已知拋物線SKIPIF1<0：SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，拋物線SKIPIF1<0上有一動(dòng)點(diǎn)SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．8 B．16 C．11 D．26【答案】C【詳解】因?yàn)閽佄锞€SKIPIF1<0：SKIPIF1<0，所以拋物線SKIPIF1<0的準(zhǔn)線為SKIPIF1<0，記拋物線SKIPIF1<0的準(zhǔn)線為SKIPIF1<0，作SKIPIF1<0于SKIPIF1<0，如圖所示：因?yàn)镾KIPIF1<0，SKIPIF1<0，所以當(dāng)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0共線時(shí)，SKIPIF1<0有最小值，最小值為SKIPIF1<0.故選：C.【典例2】（2023春·甘肅武威·高二武威第六中學(xué)?？计谥校㏒KIPIF1<0是拋物線SKIPIF1<0的焦點(diǎn)，點(diǎn)SKIPIF1<0，SKIPIF1<0為拋物線上一點(diǎn)，SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0，則SKIPIF1<0的最小值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．3 D．SKIPIF1<0【答案】C【詳解】由題設(shè)，拋物線焦點(diǎn)SKIPIF1<0，準(zhǔn)線為SKIPIF1<0，故SKIPIF1<0，如上圖：SKIPIF1<0，僅當(dāng)SKIPIF1<0共線且SKIPIF1<0在SKIPIF1<0兩點(diǎn)之間時(shí)等號(hào)成立.故選：C【典例3】（2023·全國(guó)·高三專題練習(xí)）已知點(diǎn)SKIPIF1<0是坐標(biāo)平面內(nèi)一定點(diǎn)，若拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，點(diǎn)SKIPIF1<0是拋物線上的一動(dòng)點(diǎn)，則SKIPIF1<0的最小值是.【答案】SKIPIF1<0/SKIPIF1<0【詳解】拋物線的準(zhǔn)線方程為SKIPIF1<0，過點(diǎn)SKIPIF1<0作SKIPIF1<0垂直準(zhǔn)線于點(diǎn)SKIPIF1<0，SKIPIF1<0顯然，當(dāng)SKIPIF1<0平行于SKIPIF1<0軸時(shí)，SKIPIF1<0取得最小值，此時(shí)SKIPIF1<0，此時(shí)SKIPIF1<0故答案為:SKIPIF1<0.【變式1】（2023秋·內(nèi)蒙古巴彥淖爾·高二?？计谀c(diǎn)SKIPIF1<0是拋物線SKIPIF1<0的焦點(diǎn)，直線SKIPIF1<0為拋物線的準(zhǔn)線，點(diǎn)SKIPIF1<0為直線SKIPIF1<0上一動(dòng)點(diǎn)，點(diǎn)SKIPIF1<0在以SKIPIF1<0為圓心，SKIPIF1<0為半徑的圓上，點(diǎn)SKIPIF1<0在拋物線SKIPIF1<0上，則SKIPIF1<0的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】如圖，過點(diǎn)P作SKIPIF1<0于點(diǎn)N，根據(jù)拋物線的定義可得：SKIPIF1<0，所以SKIPIF1<0，而SKIPIF1<0所以SKIPIF1<0.當(dāng)且僅當(dāng)點(diǎn)Q、點(diǎn)N、點(diǎn)M在同一條直線上時(shí)等號(hào)成立，所以SKIPIF1<0有最大值1.故選：B【變式2】（2023秋·高二單元測(cè)試）已知拋物線SKIPIF1<0的焦點(diǎn)為F，點(diǎn)M（3，6），點(diǎn)Q在拋物線上，則SKIPIF1<0的最小值為．【答案】SKIPIF1<0【詳解】拋物線SKIPIF1<0的準(zhǔn)線方程為SKIPIF1<0，過SKIPIF1<0作準(zhǔn)線SKIPIF1<0的垂線，垂足為SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0.當(dāng)且僅當(dāng)SKIPIF1<0與準(zhǔn)線垂直時(shí)，取等號(hào).所以SKIPIF1<0的最小值為SKIPIF1<0.

故答案為：SKIPIF1<0.題型05根據(jù)拋物線方程求焦點(diǎn)和準(zhǔn)線【典例1】2．（2023春·四川·高二統(tǒng)考期末）拋物線SKIPIF1<0的焦點(diǎn)坐標(biāo)是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】由SKIPIF1<0得SKIPIF1<0，故焦點(diǎn)為SKIPIF1<0，故選：B【典例2】（2023春·上海浦東新·高二統(tǒng)考期末）拋物線SKIPIF1<0的準(zhǔn)線方程是．【答案】SKIPIF1<0【詳解】因?yàn)閽佄锞€的方程為SKIPIF1<0，所以拋物線SKIPIF1<0的準(zhǔn)線方程是SKIPIF1<0.故答案為：SKIPIF1<0.【變式1】（2023·青海西寧·統(tǒng)考二模）已知函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的圖像過定點(diǎn)A，若拋物線SKIPIF1<0也過點(diǎn)A，則拋物線的準(zhǔn)線方程為.【答案】x=-1【詳解】因?yàn)楹瘮?shù)SKIPIF1<0經(jīng)過定點(diǎn)SKIPIF1<0，所以函數(shù)SKIPIF1<0經(jīng)過定點(diǎn)SKIPIF1<0，將它代入拋物線方程得SKIPIF1<0，解得SKIPIF1<0，所以其準(zhǔn)線方程為SKIPIF1<0；故答案為：SKIPIF1<0.題型06拋物線的焦半徑公式【典例1】（2023春·廣東廣州·高二統(tǒng)考期末）已知拋物線SKIPIF1<0上的點(diǎn)SKIPIF1<0到其焦點(diǎn)的距離為SKIPIF1<0，則點(diǎn)SKIPIF1<0的橫坐標(biāo)是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】設(shè)點(diǎn)SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0，拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0，該拋物線的準(zhǔn)線方程為SKIPIF1<0，因?yàn)閽佄锞€SKIPIF1<0上的點(diǎn)SKIPIF1<0到其焦點(diǎn)的距離為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0.故選：C.【典例2】（多選）（2023秋·廣西河池·高二統(tǒng)考期末）已知拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，點(diǎn)SKIPIF1<0在拋物線SKIPIF1<0上，若SKIPIF1<0為坐標(biāo)原點(diǎn)，則（

）A．點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BD【詳解】由題可知SKIPIF1<0，因?yàn)辄c(diǎn)SKIPIF1<0在拋物線SKIPIF1<0上，且SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，故選：BD.【變式1】（2023·安徽滁州·安徽省定遠(yuǎn)中學(xué)?？级＃┮阎猄KIPIF1<0為拋物線SKIPIF1<0上一點(diǎn)，點(diǎn)SKIPIF1<0到SKIPIF1<0的焦點(diǎn)的距離為SKIPIF1<0，則SKIPIF1<0的焦點(diǎn)坐標(biāo)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】由題意可知，SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0又知拋物線SKIPIF1<0的準(zhǔn)線方程為SKIPIF1<0，根據(jù)拋物線的定義可知，SKIPIF1<0，整理得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的焦點(diǎn)坐標(biāo)為SKIPIF1<0，故選：C．【變式2】（2023春·四川宜賓·高二四川省宜賓市第四中學(xué)校校考期末）拋物線SKIPIF1<0上的點(diǎn)SKIPIF1<0到焦點(diǎn)SKIPIF1<0的距離為SKIPIF1<0，則點(diǎn)SKIPIF1<0的縱坐標(biāo)為.【答案】1【詳解】拋物線SKIPIF1<0，SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0，依題意可知，SKIPIF1<0，得SKIPIF1<0，故答案為：SKIPIF1<0題型07求拋物線方程【典例1】（2023春·四川南充·高二四川省南充高級(jí)中學(xué)校考期中）準(zhǔn)線方程為SKIPIF1<0的拋物線的標(biāo)準(zhǔn)方程是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】根據(jù)題意，拋物線的準(zhǔn)線方程為SKIPIF1<0，即其焦點(diǎn)在SKIPIF1<0軸負(fù)半軸上，且SKIPIF1<0，得SKIPIF1<0，故其標(biāo)準(zhǔn)方程為：SKIPIF1<0.故選：D.【典例2】（2023春·內(nèi)蒙古呼倫貝爾·高二校考階段練習(xí)）經(jīng)過點(diǎn)SKIPIF1<0的拋物線的標(biāo)準(zhǔn)方程是（

）A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】C【詳解】設(shè)拋物線的方程為SKIPIF1<0或SKIPIF1<0，將點(diǎn)SKIPIF1<0代入，可得SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，故拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0或SKIPIF1<0，故選：C【典例3】（2023·全國(guó)·高三專題練習(xí)）已知拋物線SKIPIF1<0同時(shí)滿足以下三個(gè)條件①SKIPIF1<0的頂點(diǎn)在坐標(biāo)原點(diǎn)；②SKIPIF1<0的對(duì)稱軸為坐標(biāo)軸；③SKIPIF1<0的焦點(diǎn)SKIPIF1<0在圓SKIPIF1<0上．則SKIPIF1<0的方程為．（寫出一個(gè)滿足題意的即可），【答案】SKIPIF1<0（答案不唯一，只需填寫SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0中的任意一個(gè)）【詳解】由已知得：拋物線SKIPIF1<0的焦點(diǎn)SKIPIF1<0在坐標(biāo)軸上；若拋物線的焦點(diǎn)在SKIPIF1<0軸上，將SKIPIF1<0代入SKIPIF1<0可得：SKIPIF1<0，SKIPIF1<0拋物線的焦點(diǎn)為SKIPIF1<0，SKIPIF1<0；當(dāng)拋物線的焦點(diǎn)為SKIPIF1<0時(shí)，拋物線的方程為SKIPIF1<0；當(dāng)拋物線的焦點(diǎn)為SKIPIF1<0時(shí)，拋物線的方程為SKIPIF1<0；若拋物線的焦點(diǎn)在SKIPIF1<0軸上，將SKIPIF1<0代入SKIPIF1<0可得：SKIPIF1<0或SKIPIF1<0，SKIPIF1<0拋物線的焦點(diǎn)為SKIPIF1<0，SKIPIF1<0；當(dāng)拋物線的焦點(diǎn)為SKIPIF1<0時(shí)，拋物線的方程為SKIPIF1<0；當(dāng)拋物線的焦點(diǎn)為SKIPIF1<0時(shí)，拋物線的方程為SKIPIF1<0；則可同時(shí)滿足三個(gè)條件的拋物線SKIPIF1<0的方程為SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0．故答案為：SKIPIF1<0（答案不唯一，只需填寫SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0中的任意一個(gè)）．【變式1】（2023·河南新鄉(xiāng)·統(tǒng)考三模）已知拋物線SKIPIF1<0的焦點(diǎn)為F，C上一點(diǎn)SKIPIF1<0滿足SKIPIF1<0，則拋物線C的方程為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】解：依題意得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0.又SKIPIF1<0，解得SKIPIF1<0，所以拋物線SKIPIF1<0的方程為SKIPIF1<0.故選：D【變式2】（2023·全國(guó)·高三專題練習(xí)）設(shè)點(diǎn)F是拋物線SKIPIF1<0的焦點(diǎn)，l是該拋物線的準(zhǔn)線，過拋物線上一點(diǎn)A作準(zhǔn)線的垂線AB，垂足為B，射線AF交準(zhǔn)線l于點(diǎn)C，若SKIPIF1<0，SKIPIF1<0，則拋物線的方程為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】解：由題意得：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0可得SKIPIF1<0，由拋物線的定義得SKIPIF1<0所以SKIPIF1<0是等邊三角形，所以SKIPIF1<0，所以拋物線的方程是SKIPIF1<0．故選：B【變式3】（2023·全國(guó)·高二專題練習(xí)）已知拋物線SKIPIF1<0上一點(diǎn)SKIPIF1<0到焦點(diǎn)SKIPIF1<0的距離SKIPIF1<0．求拋物線SKIPIF1<0的方程；【答案】SKIPIF1<0【詳解】因?yàn)閽佄锞€SKIPIF1<0上一點(diǎn)SKIPIF1<0到焦點(diǎn)SKIPIF1<0的距離SKIPIF1<0，所以拋物線的定義得SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0．所以拋物線的方程為SKIPIF1<0；題型08拋物線的實(shí)際問題【典例1】（2023·全國(guó)·高二專題練習(xí)）清代青花瓷蓋碗是中國(guó)傳統(tǒng)茶文化的器物載體，具有“溫潤(rùn)”“淡遠(yuǎn)”“清新”的特征．如圖，已知碗體和碗蓋的內(nèi)部均近似為拋物線形狀，碗蓋深為SKIPIF1<0，碗蓋口直徑為SKIPIF1<0，碗體口直徑為SKIPIF1<0，碗體深SKIPIF1<0，則蓋上碗蓋后，碗蓋內(nèi)部最高點(diǎn)到碗底的垂直距離為（碗和碗蓋的厚度忽略不計(jì)）（

）

A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】以碗體的最低點(diǎn)為原點(diǎn)，向上方向?yàn)镾KIPIF1<0軸，建立直角坐標(biāo)系，如圖所示．

設(shè)碗體的拋物線方程為SKIPIF1<0（SKIPIF1<0），將點(diǎn)SKIPIF1<0代入，得SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0，設(shè)蓋上碗蓋后，碗蓋內(nèi)部最高點(diǎn)到碗底的垂直距離為SKIPIF1<0SKIPIF1<0，則兩拋物線在第一象限的交點(diǎn)為SKIPIF1<0，代入到SKIPIF1<0，解得SKIPIF1<0，解得SKIPIF1<0．故選：C【典例2】（2023春·廣東韶關(guān)·高二?？茧A段練習(xí)）有一個(gè)隧道內(nèi)設(shè)雙行線公路，其截面由一長(zhǎng)方形和拋物線構(gòu)成，如圖所示.為了保證安全，要求行駛車輛頂部（設(shè)為平頂）與隧道頂部在豎直方向上的高度之差至少為0.7m，若行車道總寬度為7.2m，則車輛通過隧道時(shí)的限制高度為m.【答案】3.8【詳解】由題意，如圖建系：則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，如圖可設(shè)，拋物線方程為SKIPIF1<0，將SKIPIF1<0代入，可得SKIPIF1<0，求得SKIPIF1<0，故拋物線方程為SKIPIF1<0，將SKIPIF1<0代入拋物線方程，可得SKIPIF1<0，SKIPIF1<0.故答案為：3.8.【變式1】（2023春·甘肅白銀·高二校考期末）圖中是拋物線形拱橋，當(dāng)水面在SKIPIF1<0時(shí)，拱頂距離水面2米，水面寬度為8米，則當(dāng)水面寬度為10米時(shí)，拱頂與水面之間的距離為（

）A．SKIPIF1<0米 B．SKIPIF1<0米 C．SKIPIF1<0米 D．SKIPIF1<0米【答案】D【詳解】以拱頂為坐標(biāo)原點(diǎn)，建立直角坐標(biāo)系，可設(shè)拱橋所在拋物線的方程為SKIPIF1<0，又拋物線過點(diǎn)SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，則拋物線的方程為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故當(dāng)水面寬度為SKIPIF1<0米時(shí)，拱頂與水面之間的距離為SKIPIF1<0米.故選：D【變式2】（2023·全國(guó)·高三專題練習(xí)）數(shù)學(xué)與建筑的結(jié)合造就建筑藝術(shù)，如圖，吉林大學(xué)的校門是一拋物線形水泥建筑物，若將校門輪廓（忽略水泥建筑的厚度）近似看成拋物線SKIPIF1<0的一部分，其焦點(diǎn)坐標(biāo)為SKIPIF1<0．校門最高點(diǎn)到地面距離約為18.2米，則校門位于地面寬度最大約為（

）A．18米 B．21米 C．24米 D．27米【答案】C【詳解】依題意知，拋物線SKIPIF1<0，即SKIPIF1<0，因?yàn)閽佄锞€的焦點(diǎn)坐標(biāo)為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以拋物線方程為SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，所以校門位于地面寬度最大約為SKIPIF1<0米.故選：C.A夯實(shí)基礎(chǔ)B能力提升C綜合素養(yǎng)A夯實(shí)基礎(chǔ)一、單選題1．（2023春·江西萍鄉(xiāng)·高二校聯(lián)考階段練習(xí)）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．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】由拋物線SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，所以拋物線的焦點(diǎn)坐標(biāo)為SKIPIF1<0，準(zhǔn)線方程為SKIPIF1<0所以該拋物線的焦點(diǎn)到其準(zhǔn)線的距離為SKIPIF1<0.故選：C.2．（2023春·河南南陽(yáng)·高二校聯(lián)考階段練習(xí)）拋物線C：SKIPIF1<0過點(diǎn)SKIPIF1<0，則C的準(zhǔn)線方程為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】拋物線C：SKIPIF1<0過點(diǎn)SKIPIF1<0，則SKIPIF1<0，解之得SKIPIF1<0，則拋物線C方程為SKIPIF1<0，則C的準(zhǔn)線方程為SKIPIF1<0故選：B3．（2023春·廣東東莞·高二校聯(lián)考階段練習(xí)）一種衛(wèi)星接收天線（如圖1），其曲面與軸截面的交線可視為拋物線的一部分（如圖2），已知該衛(wèi)星接收天線的口徑SKIPIF1<0米，深度SKIPIF1<0米，信號(hào)處理中心SKIPIF1<0位于焦點(diǎn)處，以頂點(diǎn)SKIPIF1<0為坐標(biāo)原點(diǎn)，建立如圖2所示的平面直角坐標(biāo)系SKIPIF1<0，則該拋物線的方程為（

）

A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】由題意，結(jié)合圖形可知，SKIPIF1<0，由于該拋物線開口向右，可設(shè)SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，于是SKIPIF1<0.故選：B4．（2023春·湖北·高二十堰一中校聯(lián)考期中）已知SKIPIF1<0的頂點(diǎn)都在拋物線SKIPIF1<0上，且SKIPIF1<0的重心為拋物線的焦點(diǎn)F，則SKIPIF1<0（

）A．3 B．6 C．9 D．12【答案】B【詳解】由題意得SKIPIF1<0,SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0點(diǎn)是SKIPIF1<0的重心，SKIPIF1<0，SKIPIF1<0，根據(jù)拋物線的定義可得SKIPIF1<0.故選：B.5．（2023春·福建泉州·高二校聯(lián)考期中）拋物線SKIPIF1<0繞其頂點(diǎn)逆時(shí)針旋轉(zhuǎn)SKIPIF1<0之后，得到的圖象正好對(duì)應(yīng)拋物線SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．SKIPIF1<0【答案】B【詳解】拋物線SKIPIF1<0即SKIPIF1<0的開口向上，將其繞頂點(diǎn)順時(shí)針方向旋轉(zhuǎn)SKIPIF1<0，得到的拋物線SKIPIF1<0，開口向右，其方程為SKIPIF1<0，則SKIPIF1<0，故選：B.6．（2023春·陜西西安·高二統(tǒng)考期末）已知拋物線SKIPIF1<0的焦點(diǎn)為F，點(diǎn)SKIPIF1<0在C上，則SKIPIF1<0（

）A．7 B．6 C．5 D．4【答案】D【詳解】點(diǎn)SKIPIF1<0在C：SKIPIF1<0上，設(shè)SKIPIF1<0，而拋物線SKIPIF1<0的焦點(diǎn)坐標(biāo)為SKIPIF1<0，故SKIPIF1<0，則SKIPIF1<0.故選：D7．（2023·河南鄭州·統(tǒng)考模擬預(yù)測(cè)）已知拋物線SKIPIF1<0，F(xiàn)為拋物線的焦點(diǎn)，P為拋物線上一點(diǎn)，過點(diǎn)P作PQ垂直于拋物線的準(zhǔn)線，垂足為Q，若SKIPIF1<0，則△PFQ的面積為（

）A．4 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】拋物線的準(zhǔn)線方程為y＝－1，焦點(diǎn)為SKIPIF1<0，設(shè)點(diǎn)P的坐標(biāo)為SKIPIF1<0，則點(diǎn)Q的坐標(biāo)為SKIPIF1<0，SKIPIF1<0，由拋物線的定義知SKIPIF1<0，因?yàn)镾KIPIF1<0，所以△PFQ為等邊三角形，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，n＝3，所以點(diǎn)P的坐標(biāo)為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0．故選：C．

8．（2023·重慶萬州·重慶市萬州第三中學(xué)?？寄M預(yù)測(cè)）過拋物線SKIPIF1<0的焦點(diǎn)SKIPIF1<0，作傾斜角為SKIPIF1<0的直線SKIPIF1<0交SKIPIF1<0于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，交SKIPIF1<0的準(zhǔn)線于點(diǎn)SKIPIF1<0，若SKIPIF1<0（SKIPIF1<0為坐標(biāo)原點(diǎn)），則線段SKIPIF1<0的長(zhǎng)度為（

）A．8 B．16 C．24 D．32【答案】D【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的方程為SKIPIF1<0，聯(lián)立SKIPIF1<0可得SKIPIF1<0，即點(diǎn)SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以直線SKIPIF1<0的方程為SKIPIF1<0，拋物線SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0，SKIPIF1<0，聯(lián)立SKIPIF1<0可得SKIPIF1<0，由韋達(dá)定理可得SKIPIF1<0，則SKIPIF1<0故選：D二、多選題9．（2023秋·江蘇鹽城·高二鹽城市伍佑中學(xué)?？计谀┫铝姓f法中，正確的有（

）A．過點(diǎn)SKIPIF1<0并且傾斜角為0°的直線方程為SKIPIF1<0B．雙曲線SKIPIF1<0的漸近線方程為SKIPIF1<0C．點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0的對(duì)稱點(diǎn)坐標(biāo)為SKIPIF1<0D．拋物線SKIPIF1<0的準(zhǔn)線方程是SKIPIF1<0【答案】BC【詳解】對(duì)A，過點(diǎn)SKIPIF1<0并且傾斜角為0°的直線方程為SKIPIF1<0，故錯(cuò)誤；對(duì)B，雙曲線SKIPIF1<0的漸近線方程為SKIPIF1<0，故正確；對(duì)C，設(shè)點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0的對(duì)稱點(diǎn)坐標(biāo)為SKIPIF1<0，則由SKIPIF1<0解得SKIPIF1<0，故正確；對(duì)D，拋物線SKIPIF1<0，SKIPIF1<0，準(zhǔn)線方程為SKIPIF1<0，故錯(cuò)誤.故選：BC10．（2023春·廣西·高二校聯(lián)考期中）已知雙曲線SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0，拋物線SKIPIF1<0的焦點(diǎn)與雙曲線C的一個(gè)焦點(diǎn)重合，點(diǎn)P是這兩條曲線的一個(gè)公共點(diǎn)，則下列說法正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0的周長(zhǎng)為16C．SKIPIF1<0的面積為SKIPIF1<0 D．SKIPIF1<0【答案】AB【詳解】由已知，雙曲線右焦點(diǎn)SKIPIF1<0，即SKIPIF1<0，故A項(xiàng)正確．且拋物線方程為SKIPIF1<0．對(duì)于B項(xiàng)，聯(lián)立雙曲線與拋物線的方程SKIPIF1<0，整理可得．SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去負(fù)值），所以SKIPIF1<0，代入SKIPIF1<0可得，SKIPIF1<0．設(shè)SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的周長(zhǎng)為16，故B項(xiàng)正確；對(duì)于C項(xiàng)，易知SKIPIF1<0，故C項(xiàng)錯(cuò)誤；對(duì)于D項(xiàng)，由余弦定理可得，SKIPIF1<0，故D項(xiàng)錯(cuò)誤．故選：AB

三、填空題11．（2023秋·高二課時(shí)練習(xí)）點(diǎn)SKIPIF1<0到拋物線SKIPIF1<0的準(zhǔn)線的距離為6，那么拋物線的方程是．【答案】SKIPIF1<0【詳解】當(dāng)SKIPIF1<0時(shí)，準(zhǔn)線SKIPIF1<0，由已知得SKIPIF1<0，所以SKIPIF1<0，所以拋物線方程為SKIPIF1<0．故答案為：SKIPIF1<0.12．（2023春·江蘇南京·高二南京市江寧高級(jí)中學(xué)校聯(lián)考期末）已知拋物線C：SKIPIF1<0的焦點(diǎn)為F，準(zhǔn)線為SKIPIF1<0，經(jīng)過點(diǎn)F的直線與拋物線C相交A，B兩點(diǎn)，SKIPIF1<0與x軸相交于點(diǎn)M，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0.【答案】4【詳解】

由題意易知SKIPIF1<0，可設(shè)SKIPIF1<0，由SKIPIF1<0，可得Q為AM中點(diǎn)，則SKIPIF1<0，又由SKIPIF1<0可得:SKIPIF1<0，即SKIPIF1<0，由題意可知直線AB、BM的斜率存在，故SKIPIF1<0，聯(lián)立拋物線與直線AB可得SKIPIF1<0所以有SKIPIF1<0由拋物線定義得SKIPIF1<0，故答案為：4四、解答題13．（2023春·四川遂寧·高二統(tǒng)考期末）分別求適合下列條件的方程：(1)長(zhǎng)軸長(zhǎng)為10，焦距為4的橢圓標(biāo)準(zhǔn)方程；(2)經(jīng)過點(diǎn)SKIPIF1<0的拋物線的標(biāo)準(zhǔn)方程.【答案】(1)SKIPIF1<0或SKIPIF1<0(2)SKIPIF1<0或SKIPIF1<0【詳解】（1）設(shè)橢圓的長(zhǎng)軸長(zhǎng)為SKIPIF1<0，焦距為SKIPIF1<0由條件可得SKIPIF1<0.所以SKIPIF1<0.所以SKIPIF1<0，當(dāng)橢圓的焦點(diǎn)在SKIPIF1<0軸上時(shí)，標(biāo)準(zhǔn)方程為SKIPIF1<0；當(dāng)橢圓的焦點(diǎn)在SKIPIF1<0軸上時(shí)，標(biāo)準(zhǔn)方程為SKIPIF1<0.（2）當(dāng)拋物線的焦點(diǎn)在SKIPIF1<0軸上時(shí)，可設(shè)所求拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0，將點(diǎn)SKIPIF1<0的坐標(biāo)代入拋物線的標(biāo)準(zhǔn)方程得SKIPIF1<0，此時(shí)，所求拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0；當(dāng)拋物線的焦點(diǎn)在SKIPIF1<0軸上時(shí)，可設(shè)所求拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0，將點(diǎn)SKIPIF1<0的坐標(biāo)代入拋物線的標(biāo)準(zhǔn)方程得SKIPIF1<0，解得SKIPIF1<0，此時(shí)，所求拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0.綜上所述，所求拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0或SKIPIF1<0.14．（2023春·四川成都·高二?？茧A段練習(xí)）動(dòng)點(diǎn)SKIPIF1<0與定點(diǎn)SKIPIF1<0的距離等于點(diǎn)P到直線SKIPIF1<0的距離，設(shè)動(dòng)點(diǎn)P的軌跡為曲線SKIPIF1<0．(1)求曲線SKIPIF1<0的方程；(2)經(jīng)過定點(diǎn)SKIPIF1<0直線SKIPIF1<0與曲線SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，且點(diǎn)M是線段AB的中點(diǎn)，求直線SKIPIF1<0的方程．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）根據(jù)拋物線的定義可知，動(dòng)點(diǎn)P的軌跡為拋物線，且該拋物線以SKIPIF1<0為焦點(diǎn)，所以SKIPIF1<0所以SKIPIF1<0，所以曲線SKIPIF1<0的方程為SKIPIF1<0.（2）若直線SKIPIF1<0垂直于SKIPIF1<0軸，則AB的中點(diǎn)在SKIPIF1<0軸上，不滿足題意，若直線SKIPIF1<0不垂直于SKIPIF1<0軸，設(shè)SKIPIF1<0，且SKIPIF1<0，因?yàn)镾KIP