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專題23圓錐曲線【考綱要求】1、掌握橢圓的定義、幾何圖形、標(biāo)準(zhǔn)方程,掌握橢圓的簡(jiǎn)單幾何性質(zhì)(范圍、對(duì)稱性、頂點(diǎn)、離心率).2、掌握雙曲線定義、幾何圖形和標(biāo)準(zhǔn)方程,知道雙曲線簡(jiǎn)單(范圍、對(duì)稱性、頂點(diǎn)、離心率、漸近線).3、掌握拋物線的定義、幾何圖形、標(biāo)準(zhǔn)方程及簡(jiǎn)單幾何性質(zhì).一、橢圓及相關(guān)問題【思維導(dǎo)圖】【考點(diǎn)總結(jié)】一、橢圓的定義及標(biāo)準(zhǔn)方程1.定義平面內(nèi)與兩個(gè)定點(diǎn)F1,F(xiàn)2的距離的和等于常數(shù)(大于|F1F2|)的點(diǎn)的軌跡叫做橢圓.這兩個(gè)定點(diǎn)叫做橢圓的焦點(diǎn),兩焦點(diǎn)間的距離叫做橢圓的焦距.集合P={M||MF1|+|MF2|=2a,|F1F2|=2c,其中a>0,c>0,且a,c為常數(shù)}。(1)若a>c,則M點(diǎn)的軌跡為橢圓。(2)若a=c,則M點(diǎn)的軌跡為線段F1F2。(3)若a<c,則M點(diǎn)不存在。2.標(biāo)準(zhǔn)方程中心在坐標(biāo)原點(diǎn),焦點(diǎn)在x軸上的橢圓的標(biāo)準(zhǔn)方程為:eq\f(x2,a2)+eq\f(y2,b2)=1(a>b>0);中心在坐標(biāo)原點(diǎn),焦點(diǎn)在y軸上的橢圓的標(biāo)準(zhǔn)方程為:eq\f(y2,a2)+eq\f(x2,b2)=1(a>b>0).二、橢圓的標(biāo)準(zhǔn)方程和幾何性質(zhì)標(biāo)準(zhǔn)方程eq\f(x2,a2)+eq\f(y2,b2)=1(a>b>0)eq\f(y2,a2)+eq\f(x2,b2)=1(a>b>0)圖形性質(zhì)范圍-a≤x≤a-b≤y≤b-b≤x≤b-a≤y≤a對(duì)稱性對(duì)稱軸:坐標(biāo)軸;對(duì)稱中心:原點(diǎn)頂點(diǎn)A1(-a,0),A2(a,0)B1(0,-b),B2(0,b)A1(0,-a),A2(0,a)B1(-b,0),B2(b,0)軸長(zhǎng)軸A1A2的長(zhǎng)為2a;短軸B1B2的長(zhǎng)為2b焦距|F1F2|=2c離心率e=eq\f(c,a)∈(0,1)a,b,c的關(guān)系c2=a2-b2二、雙曲線及相關(guān)問題【思維導(dǎo)圖】【考點(diǎn)總結(jié)】一、雙曲線的定義及標(biāo)準(zhǔn)方程1.定義在平面內(nèi)到兩定點(diǎn)F1,F(xiàn)2的距離的差的絕對(duì)值等于常數(shù)(小于|F1F2|且大于零)的點(diǎn)的軌跡(或集合)叫做雙曲線.定點(diǎn)F1,F(xiàn)2叫做雙曲線的焦點(diǎn),兩焦點(diǎn)間的距離叫做焦距.集合P={M|||MF1|-|MF2||=2a,|F1F2|=2c,其中a、c為常數(shù)且a>0,c>0}。(1)當(dāng)a<c時(shí),M點(diǎn)的軌跡是雙曲線。(2)當(dāng)a=c時(shí),M點(diǎn)的軌跡是兩條射線。(3)當(dāng)a>c時(shí),M點(diǎn)不存在。2.標(biāo)準(zhǔn)方程中心在坐標(biāo)原點(diǎn),焦點(diǎn)在x軸上的雙曲線的標(biāo)準(zhǔn)方程為eq\f(x2,a2)-eq\f(y2,b2)=1(a>0,b>0);中心在坐標(biāo)原點(diǎn),焦點(diǎn)在y軸上的雙曲線的標(biāo)準(zhǔn)方程為eq\f(y2,a2)-eq\f(x2,b2)=1(a>0,b>0).二、雙曲線的標(biāo)準(zhǔn)方程和幾何性質(zhì)標(biāo)準(zhǔn)方程eq\f(x2,a2)-eq\f(y2,b2)=1(a>0,b>0)eq\f(y2,a2)-eq\f(x2,b2)=1(a>0,b>0)圖形性質(zhì)范圍x≥a或x≤-a,y∈Rx∈R,y≤-a或y≥a對(duì)稱性對(duì)稱軸:坐標(biāo)軸對(duì)稱中心:原點(diǎn)對(duì)稱軸:坐標(biāo)軸對(duì)稱中心:原點(diǎn)頂點(diǎn)頂點(diǎn)坐標(biāo):A1(-a,0),A2(a,0)頂點(diǎn)坐標(biāo):A1(0,-a),A2(0,a)漸近線y=±eq\f(b,a)xy=±eq\f(a,b)x離心率e=eq\f(c,a),e∈(1,+∞),其中c=eq\r(a2+b2)性質(zhì)實(shí)虛軸線段A1A2叫做雙曲線的實(shí)軸,它的長(zhǎng)|A1A2|=2a;線段B1B2叫做雙曲線的虛軸,它的長(zhǎng)|B1B2|=2b;a叫做雙曲線的實(shí)半軸長(zhǎng),b叫做雙曲線的虛半軸長(zhǎng)三、拋物線及相關(guān)問題【思維導(dǎo)圖】【考點(diǎn)總結(jié)】一、拋物線的定義及標(biāo)準(zhǔn)方程1.定義平面內(nèi)與一個(gè)定點(diǎn)F和一條定直線l(F?l)的距離相等的點(diǎn)的軌跡叫做拋物線,點(diǎn)F叫做拋物線的焦點(diǎn),直線l叫做拋物線的準(zhǔn)線。2.標(biāo)準(zhǔn)方程頂點(diǎn)在坐標(biāo)原點(diǎn),焦點(diǎn)在x軸正半軸上的拋物線的標(biāo)準(zhǔn)方程為:y2=2px(p>0);頂點(diǎn)在坐標(biāo)原點(diǎn),焦點(diǎn)在x軸負(fù)半軸上的拋物線的標(biāo)準(zhǔn)方程為:y2=-2px(p>0);頂點(diǎn)在坐標(biāo)原點(diǎn),焦點(diǎn)在y軸正半軸上的拋物線的標(biāo)準(zhǔn)方程為:x2=2py(p>0);頂點(diǎn)在坐標(biāo)原點(diǎn),焦點(diǎn)在y軸負(fù)半軸上的拋物線的標(biāo)準(zhǔn)方程為:x2=-2py(p>0).二、拋物線的標(biāo)準(zhǔn)方程與幾何性質(zhì)標(biāo)準(zhǔn)方程y2=2px(p>0)y2=-2px(p>0)x2=2py(p>0)x2=-2py(p>0)p的幾何意義:焦點(diǎn)F到準(zhǔn)線l的距離圖形頂點(diǎn)O(0,0)對(duì)稱軸y=0x=0焦點(diǎn)Feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(p,2),0))Feq\b\lc\(\rc\)(\a\vs4\al\co1(-\f(p,2),0))Feq\b\lc\(\rc\)(\a\vs4\al\co1(0,\f(p,2)))Feq\b\lc\(\rc\)(\a\vs4\al\co1(0,-\f(p,2)))離心率e=1準(zhǔn)線方程x=-eq\f(p,2)x=eq\f(p,2)y=-eq\f(p,2)y=eq\f(p,2)范圍x≥0,y∈Rx≤0,y∈Ry≥0,x∈Ry≤0,x∈R開口方向向右向左向上向下焦半徑|PF|=x0+eq\f(p,2)|PF|=-x0+eq\f(p,2)|PF|=y(tǒng)0+eq\f(p,2)|PF|=-y0+eq\f(p,2)【題型匯編】題型一:橢圓題型二:雙曲線題型三:拋物線【題型講解】題型一:橢圓一、單選題1.(2022·全國(guó)·一模(理))已知橢圓C:SKIPIF1<0上的動(dòng)點(diǎn)P到右焦點(diǎn)距離的最小值為SKIPIF1<0,則SKIPIF1<0(

)A.1 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】根據(jù)橢圓的性質(zhì)可得橢圓上的點(diǎn)到右焦點(diǎn)距離最小值為SKIPIF1<0,即可求出SKIPIF1<0,再根據(jù)SKIPIF1<0,即可得解;【詳解】解:根據(jù)橢圓的性質(zhì),橢圓上的點(diǎn)到右焦點(diǎn)距離最小值為SKIPIF1<0,即SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0,由SKIPIF1<0,所以SKIPIF1<0;故選:A2.(2022·山西大附中三模(文))已知橢圓C:SKIPIF1<0的右焦點(diǎn)為SKIPIF1<0,右頂點(diǎn)為A,O為坐標(biāo)原點(diǎn),過OA的中點(diǎn)且與坐標(biāo)軸垂直的直線交橢圓C于M,N兩點(diǎn),若四邊形OMAN是正方形,則C的方程為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】待定系數(shù)法去求橢圓C的方程【詳解】由橢圓方程可知SKIPIF1<0,由四邊形OMAN是正方形可知SKIPIF1<0,又點(diǎn)M在橢圓C上,則有SKIPIF1<0,解得SKIPIF1<0,又橢圓C的右焦點(diǎn)為SKIPIF1<0,則SKIPIF1<0,結(jié)合橢圓中SKIPIF1<0,解得SKIPIF1<0,SKIPIF1<0,則橢圓C的方程為SKIPIF1<0.故選:A3.(2022·湖南湘潭·三模)橢圓SKIPIF1<0的左?右焦點(diǎn)分別為F1,F(xiàn)2,過點(diǎn)F1的直線l與E交于A,B兩點(diǎn),若△ABF2的周長(zhǎng)為12,則E的離心率為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】由橢圓的定義,求得SKIPIF1<0,再由SKIPIF1<0,求得SKIPIF1<0的值,結(jié)合離心率的定義,即可求解.【詳解】因?yàn)镾KIPIF1<0的周長(zhǎng)為SKIPIF1<0,根據(jù)橢圓的定義可得SKIPIF1<0,解得SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,則橢圓SKIPIF1<0的離心率為SKIPIF1<0.故選:A.4.(2022·寧夏·銀川一中二模(文))橢圓SKIPIF1<0的一個(gè)焦點(diǎn)坐標(biāo)為SKIPIF1<0,則實(shí)數(shù)m的值為(

)A.2 B.4 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】由焦點(diǎn)坐標(biāo)得到SKIPIF1<0,求解即可.【詳解】根據(jù)焦點(diǎn)坐標(biāo)可知,橢圓焦點(diǎn)在y軸上,所以有SKIPIF1<0,解得SKIPIF1<0.故選:C.5.(2022·陜西西安·二模(文))已知橢圓SKIPIF1<0的兩焦點(diǎn)為SKIPIF1<0,以SKIPIF1<0為邊作正三角形,若橢圓恰好平分正三角形的另兩條邊,則橢圓的離心率為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】橢圓的正三角形另兩邊的交點(diǎn)分別為SKIPIF1<0,易得SKIPIF1<0,SKIPIF1<0,由此建立SKIPIF1<0的齊次式,進(jìn)而可的結(jié)果.【詳解】解:由題意得:設(shè)橢圓的正三角形另兩邊的交點(diǎn)分別為SKIPIF1<0,易得SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0故選:A6.(2022·湖南·長(zhǎng)沙市明德中學(xué)二模)已知SKIPIF1<0分別是橢圓SKIPIF1<0的左、右焦點(diǎn),點(diǎn)SKIPIF1<0,點(diǎn)SKIPIF1<0在橢圓SKIPIF1<0上,SKIPIF1<0,SKIPIF1<0分別是SKIPIF1<0的中點(diǎn),且SKIPIF1<0的周長(zhǎng)為SKIPIF1<0,則橢圓SKIPIF1<0的方程為(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】因?yàn)镾KIPIF1<0,所以SKIPIF1<0三點(diǎn)共線,且SKIPIF1<0,根據(jù)橢圓的定義求得SKIPIF1<0,設(shè)SKIPIF1<0,根據(jù)SKIPIF1<0,求得SKIPIF1<0,代入橢圓的方程,求得SKIPIF1<0的值,即可求解.【詳解】因?yàn)镾KIPIF1<0,所以SKIPIF1<0三點(diǎn)共線,且SKIPIF1<0,因?yàn)镾KIPIF1<0分別為SKIPIF1<0和SKIPIF1<0的中點(diǎn),所以SKIPIF1<0,所以SKIPIF1<0,設(shè)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,由SKIPIF1<0,可得SKIPIF1<0,求得SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,因?yàn)辄c(diǎn)SKIPIF1<0在橢圓SKIPIF1<0上,所以SKIPIF1<0,求得SKIPIF1<0,SKIPIF1<0,所以橢圓SKIPIF1<0的方程為SKIPIF1<0.故選:B.二、多選題7.(2022·江蘇江蘇·一模)若橢圓SKIPIF1<0的左,右焦點(diǎn)分別為SKIPIF1<0,則下列SKIPIF1<0的值,能使以SKIPIF1<0為直徑的圓與橢圓SKIPIF1<0有公共點(diǎn)的有(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】ABC【解析】【分析】依題意可得SKIPIF1<0,再根據(jù)SKIPIF1<0,即可取出SKIPIF1<0的取值范圍,即可得解;【詳解】解:以SKIPIF1<0為直徑的圓的方程為SKIPIF1<0,因?yàn)閳ASKIPIF1<0與橢圓SKIPIF1<0有公共點(diǎn),所以SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,滿足條件的有A、B、C;故選:ABC2.(2022·湖北·黃岡中學(xué)二模)已知點(diǎn)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0是橢圓SKIPIF1<0上的動(dòng)點(diǎn),當(dāng)SKIPIF1<0取下列哪些值時(shí),可以使SKIPIF1<0(

)A.3 B.6 C.9 D.12【答案】ABC【解析】【分析】設(shè)SKIPIF1<0,利用SKIPIF1<0求得SKIPIF1<0的最大值和最小值即可得.【詳解】設(shè)SKIPIF1<0,且SKIPIF1<0.因?yàn)镾KIPIF1<0,將SKIPIF1<0點(diǎn)坐標(biāo)代入橢圓,得SKIPIF1<0,所以SKIPIF1<0代入上式可得SKIPIF1<0SKIPIF1<0.所以SKIPIF1<0,SKIPIF1<0.對(duì)照選項(xiàng)SKIPIF1<0可以取ABC.故選:ABC.三、解答題1.(2022·北京·北大附中三模)已知橢圓SKIPIF1<0經(jīng)過點(diǎn)SKIPIF1<0.(1)求橢圓SKIPIF1<0的方程及其離心率;(2)若SKIPIF1<0為橢圓SKIPIF1<0上第一象限的點(diǎn),直線SKIPIF1<0交SKIPIF1<0軸于點(diǎn)SKIPIF1<0,直線SKIPIF1<0交SKIPIF1<0軸于點(diǎn)SKIPIF1<0,且有SKIPIF1<0,求點(diǎn)SKIPIF1<0的坐標(biāo).【答案】(1)SKIPIF1<0,離心率為SKIPIF1<0;(2)SKIPIF1<0【解析】【分析】(1)由題意可得SKIPIF1<0,繼而求出SKIPIF1<0,即可得方程和離心率;(2)設(shè)SKIPIF1<0,則SKIPIF1<0,又由SKIPIF1<0可得SKIPIF1<0,繼而得到SKIPIF1<0,聯(lián)立即可解得SKIPIF1<0,SKIPIF1<0的值.(1)依題知:SKIPIF1<0,所以SKIPIF1<0.所以橢圓方程為SKIPIF1<0,離心率SKIPIF1<0.(2)如圖:設(shè)SKIPIF1<0,第一象限有SKIPIF1<0,SKIPIF1<0①;由SKIPIF1<0得:SKIPIF1<0,又SKIPIF1<0,SKIPIF1<0,因此SKIPIF1<0②,聯(lián)立①②解得SKIPIF1<0,故SKIPIF1<0.2.(2022·海南海口·二模)已知橢圓SKIPIF1<0的離心率為SKIPIF1<0,且經(jīng)過點(diǎn)SKIPIF1<0.(1)求C的方程;(2)動(dòng)直線l與圓SKIPIF1<0相切,與C交于M,N兩點(diǎn),求O到線段MN的中垂線的最大距離.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)首先根據(jù)題意列出方程組SKIPIF1<0,再解方程組即可.(2)當(dāng)SKIPIF1<0的斜率不存在時(shí),SKIPIF1<0到中垂線的距離為0.當(dāng)SKIPIF1<0的斜率存在時(shí),設(shè)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.根據(jù)直線與圓相切得到SKIPIF1<0,求出中垂線得到SKIPIF1<0到中垂線的距離為SKIPIF1<0,再利用基本不等式即可得到答案.(1)由題知:SKIPIF1<0,解得SKIPIF1<0.所以SKIPIF1<0的方程為SKIPIF1<0.(2)當(dāng)SKIPIF1<0的斜率不存在時(shí),線段MN的中垂線為SKIPIF1<0軸,此時(shí)SKIPIF1<0到中垂線的距離為0.當(dāng)SKIPIF1<0的斜率存在時(shí),設(shè)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.因?yàn)镾KIPIF1<0與圓SKIPIF1<0相切,則SKIPIF1<0到SKIPIF1<0的距離為SKIPIF1<0,所以SKIPIF1<0.聯(lián)立方程SKIPIF1<0,得SKIPIF1<0,則SKIPIF1<0,可得SKIPIF1<0的中點(diǎn)為SKIPIF1<0.則MN的中垂線方程為SKIPIF1<0,即SKIPIF1<0.因此SKIPIF1<0到中垂線的距離為SKIPIF1<0(當(dāng)且僅當(dāng)SKIPIF1<0,SKIPIF1<0時(shí)等號(hào)成立).綜上所述,SKIPIF1<0到線段MN的中垂線的最大距離為SKIPIF1<0.題型二:雙曲線一、單選題1.(2022·浙江·三模)雙曲線SKIPIF1<0的實(shí)軸長(zhǎng)度是(

)A.1 B.2 C.SKIPIF1<0 D.4【答案】D【解析】【分析】由雙曲線的幾何性質(zhì)即可得出答案.【詳解】SKIPIF1<0的SKIPIF1<0,所以SKIPIF1<0.故雙曲線SKIPIF1<0的實(shí)軸長(zhǎng)度是SKIPIF1<0.故選:D.2.(2022·安徽省舒城中學(xué)三模(理))若雙曲線SKIPIF1<0(a>0,b>0)的離心率為2,則其兩條漸近線所成的銳角為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】根據(jù)離心率可求出兩條漸近線的傾斜角,從而解出.【詳解】因?yàn)殡p曲線SKIPIF1<0的漸近線方程為SKIPIF1<0,而SKIPIF1<0,所以SKIPIF1<0,故兩條漸近線中一條的傾斜角為SKIPIF1<0,一條的傾斜角為SKIPIF1<0,它們所成的銳角為SKIPIF1<0.故選:A.3.(2022·黑龍江·哈九中三模(文))雙曲線SKIPIF1<0的漸近線方程為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】由雙曲線方程可判斷雙曲線的焦點(diǎn)位置并同時(shí)求出SKIPIF1<0,SKIPIF1<0,由此可求其漸近線方程.【詳解】由雙曲線SKIPIF1<0得SKIPIF1<0,所以漸近線方程為SKIPIF1<0,故選:B4.(2022·北京·二模)已知雙曲線SKIPIF1<0的一條漸近線方程為SKIPIF1<0,則C的離心率為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.2 D.SKIPIF1<0【答案】A【解析】【分析】根據(jù)已知漸近線確定雙曲線參數(shù),進(jìn)而求其離心率.【詳解】由題設(shè)雙曲線漸近線為SKIPIF1<0,而其中一條為SKIPIF1<0,所以SKIPIF1<0,則SKIPIF1<0,故C的離心率為SKIPIF1<0.故選:A5.(2022·北京房山·二模)雙曲線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【解析】【分析】根據(jù)雙曲線焦點(diǎn)坐標(biāo)公式求解即可【詳解】雙曲線SKIPIF1<0的焦點(diǎn)在SKIPIF1<0軸上,坐標(biāo)為SKIPIF1<0,即SKIPIF1<0故選:C6.(2022·山東煙臺(tái)·三模)過雙曲線SKIPIF1<0:SKIPIF1<0(SKIPIF1<0,SKIPIF1<0)的焦點(diǎn)且斜率不為0的直線交SKIPIF1<0于A,SKIPIF1<0兩點(diǎn),SKIPIF1<0為SKIPIF1<0中點(diǎn),若SKIPIF1<0,則SKIPIF1<0的離心率為(

)A.SKIPIF1<0 B.2 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】先設(shè)出直線AB的方程,并與雙曲線SKIPIF1<0的方程聯(lián)立,利用設(shè)而不求的方法及條件SKIPIF1<0得到關(guān)于SKIPIF1<0的關(guān)系,進(jìn)而求得雙曲線SKIPIF1<0的離心率【詳解】不妨設(shè)過雙曲線SKIPIF1<0的焦點(diǎn)且斜率不為0的直線為SKIPIF1<0,令SKIPIF1<0由SKIPIF1<0,整理得SKIPIF1<0則SKIPIF1<0,SKIPIF1<0則SKIPIF1<0,由SKIPIF1<0,可得SKIPIF1<0則有SKIPIF1<0,即SKIPIF1<0,則雙曲線SKIPIF1<0的離心率SKIPIF1<0故選:D二、多選題1.(2022·河北唐山·三模)已知SKIPIF1<0為雙曲線SKIPIF1<0的兩個(gè)焦點(diǎn),SKIPIF1<0為雙曲線SKIPIF1<0上任意一點(diǎn),則(

)A.SKIPIF1<0 B.雙曲線SKIPIF1<0的漸近線方程為SKIPIF1<0C.雙曲線SKIPIF1<0的離心率為SKIPIF1<0 D.SKIPIF1<0【答案】CD【解析】【分析】對(duì)于A,用定義即可判斷,對(duì)于B,根據(jù)焦點(diǎn)位置即可判斷,對(duì)于C,直接計(jì)算即可,對(duì)于D,因?yàn)镾KIPIF1<0為SKIPIF1<0的中點(diǎn),所以SKIPIF1<0,設(shè)SKIPIF1<0可求出SKIPIF1<0的取值范圍,即可判斷【詳解】雙曲線SKIPIF1<0:SKIPIF1<0焦點(diǎn)在SKIPIF1<0軸上,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0對(duì)于A選項(xiàng),SKIPIF1<0,而SKIPIF1<0點(diǎn)在哪支上并不確定,故A錯(cuò)誤對(duì)于B選項(xiàng),焦點(diǎn)在SKIPIF1<0軸上的雙曲線漸近線方程為SKIPIF1<0,故B錯(cuò)誤對(duì)于C選項(xiàng),SKIPIF1<0,故C正確對(duì)于D選項(xiàng),設(shè)SKIPIF1<0,則SKIPIF1<0(SKIPIF1<0時(shí)取等號(hào))因?yàn)镾KIPIF1<0為SKIPIF1<0的中點(diǎn),所以SKIPIF1<0,故D正確故選:CD三、解答題1.(2022·河北秦皇島·二模)已知雙曲線SKIPIF1<0的左?右焦點(diǎn)分別為SKIPIF1<0,SKIPIF1<0,虛軸長(zhǎng)為SKIPIF1<0,離心率為SKIPIF1<0,過SKIPIF1<0的直線SKIPIF1<0與雙曲線SKIPIF1<0的右支交于SKIPIF1<0,SKIPIF1<0兩點(diǎn).(1)求雙曲線SKIPIF1<0的方程;(2)已知SKIPIF1<0,若SKIPIF1<0的外心SKIPIF1<0的橫坐標(biāo)為0,求直線SKIPIF1<0的方程.【答案】(1)SKIPIF1<0(2)SKIPIF1<0或SKIPIF1<0【解析】【分析】(1)根據(jù)虛軸長(zhǎng)為SKIPIF1<0,離心率為SKIPIF1<0,由SKIPIF1<0求解;(2)當(dāng)直線SKIPIF1<0的斜率不存在時(shí),直線SKIPIF1<0的方程為SKIPIF1<0,根據(jù)SKIPIF1<0外接圓的圓心SKIPIF1<0的橫坐標(biāo)為0,得到SKIPIF1<0判斷.當(dāng)直線SKIPIF1<0的斜率存在時(shí),設(shè)直線SKIPIF1<0的方程為SKIPIF1<0,與雙曲線方程聯(lián)立,根據(jù)直線SKIPIF1<0與雙曲線SKIPIF1<0的右支交于SKIPIF1<0,SKIPIF1<0兩點(diǎn),求得k的范圍,設(shè)線段SKIPIF1<0的中點(diǎn)為M,利用弦長(zhǎng)公式和SKIPIF1<0求解.(1)由題知SKIPIF1<0因?yàn)镾KIPIF1<0,所以SKIPIF1<0,故雙曲線SKIPIF1<0的方程為SKIPIF1<0.(2)由(1)知SKIPIF1<0.當(dāng)直線SKIPIF1<0的斜率不存在時(shí),直線SKIPIF1<0的方程為SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0.因?yàn)镾KIPIF1<0為等腰三角形,且SKIPIF1<0外接圓的圓心SKIPIF1<0的橫坐標(biāo)為0,所以SKIPIF1<0.因?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,故此時(shí)不合題意.當(dāng)直線SKIPIF1<0的斜率存在時(shí),設(shè)直線SKIPIF1<0的方程為SKIPIF1<0,聯(lián)立方程組SKIPIF1<0得SKIPIF1<0,由SKIPIF1<0解得SKIPIF1<0,即SKIPIF1<0或SKIPIF1<0.設(shè)SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,因?yàn)镾KIPIF1<0,所以線段SKIPIF1<0的中點(diǎn)為SKIPIF1<0,且SKIPIF1<0.設(shè)SKIPIF1<0,因?yàn)镾KIPIF1<0在線段SKIPIF1<0的垂直平分線上,所以SKIPIF1<0,得SKIPIF1<0,即SKIPIF1<0,故SKIPIF1<0.因?yàn)镾KIPIF1<0,且SKIPIF1<0,所以SKIPIF1<0,化簡(jiǎn)得SKIPIF1<0,得SKIPIF1<0或SKIPIF1<0(舍去),所以直線SKIPIF1<0的方程為SKIPIF1<0,即直線SKIPIF1<0的方程為SKIPIF1<0或SKIPIF1<0.2.(2022·寧夏·銀川一中二模(理))已知雙曲線SKIPIF1<0的離心率等于SKIPIF1<0,且點(diǎn)SKIPIF1<0在雙曲線上.(1)求雙曲線的方程;(2)若雙曲線的左頂點(diǎn)為SKIPIF1<0,右焦點(diǎn)為SKIPIF1<0,P為雙曲線右支上任意一點(diǎn),求SKIPIF1<0的最小值.【答案】(1)SKIPIF1<0(2)-4【解析】【分析】(1)直接由離心率和點(diǎn)代入雙曲線求得SKIPIF1<0即可;(2)先表示出SKIPIF1<0,再通過點(diǎn)P橫坐標(biāo)的范圍求出最小值.(1)依題SKIPIF1<0又SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,故雙曲線的方程為SKIPIF1<0.(2)由已知得SKIPIF1<0,SKIPIF1<0,設(shè)SKIPIF1<0,于是SKIPIF1<0,SKIPIF1<0,因此SKIPIF1<0,由于SKIPIF1<0,所以當(dāng)SKIPIF1<0時(shí),SKIPIF1<0取得最小值,為SKIPIF1<0.題型三:拋物線一、單選題1.(2022·湖北十堰·三模)下列四個(gè)拋物線中,開口朝左的是(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】直接由拋物線的標(biāo)準(zhǔn)方程判斷開口方向即可.【詳解】拋物線SKIPIF1<0的開口朝右,拋物線SKIPIF1<0的開口朝下,拋物線SKIPIF1<0的開口朝左,拋物線SKIPIF1<0的開口朝上.故選:C.2.(2022·廣東惠州·一模)若拋物線SKIPIF1<0(SKIPIF1<0)上一點(diǎn)P(2,SKIPIF1<0)到其焦點(diǎn)的距離為4,則拋物線的標(biāo)準(zhǔn)方程為(

)A.y2=2x B.y2=4x C.y2=6x D.y2=8x【答案】D【解析】【分析】由拋物線的定義可解答.【詳解】拋物線SKIPIF1<0上一點(diǎn)SKIPIF1<0到焦點(diǎn)的距離等于到其準(zhǔn)線的距離,即為4,∴SKIPIF1<0,解得SKIPIF1<0,∴拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0.故選:D.3.(2022·陜西渭南·二模(理))拋物線SKIPIF1<0的焦點(diǎn)為F,點(diǎn)P是C上一點(diǎn),若SKIPIF1<0,則點(diǎn)P到y(tǒng)軸的距離為(

)A.2 B.3 C.4 D.5【答案】C【解析】【分析】設(shè)出SKIPIF1<0,由拋物線定義得到方程,求出SKIPIF1<0,從而得到答案.【詳解】設(shè)SKIPIF1<0,由拋物線定義知:SKIPIF1<0,所以SKIPIF1<0,即點(diǎn)P到y(tǒng)軸的距離為4故選:C4.(2022·江西九江·二模)已知點(diǎn)M為拋物線SKIPIF1<0上的動(dòng)點(diǎn),過點(diǎn)M向圓SKIPIF1<0引切線,切點(diǎn)分別為P,Q,則SKIPIF1<0的最小值為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.1【答案】A【解析】【分析】由四邊形的面積可知SKIPIF1<0,即可求解.【詳解】如圖,圓心SKIPIF1<0為拋物線的焦點(diǎn)SKIPIF1<0,四邊形SKIPIF1<0的面積SKIPIF1<0,∴SKIPIF1<0,∴當(dāng)SKIPIF1<0最小時(shí),即點(diǎn)M到準(zhǔn)線的距離最小值為2,∴SKIPIF1<0,故選:SKIPIF1<0.5.(2022·安徽馬鞍山·一模(理))已知拋物線SKIPIF1<0過點(diǎn)SKIPIF1<0,則其準(zhǔn)線方程為(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】根據(jù)拋物準(zhǔn)線方程定義即可求解.【詳解】拋物線SKIPIF1<0過點(diǎn)SKIPIF1<0,則SKIPIF1<0所以SKIPIF1<0由拋物線的標(biāo)準(zhǔn)方程可得,拋物線的焦點(diǎn)位于SKIPIF1<0軸負(fù)半軸,準(zhǔn)線方程為SKIPIF1<0.故選:D6.(2022·重慶·一模)已知拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0,準(zhǔn)線為SKIPIF1<0,點(diǎn)SKIPIF1<0在SKIPIF1<0上,直線SKIPIF1<0與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0,且SKIPIF1<0,則點(diǎn)SKIPIF1<0到準(zhǔn)線SKIPIF1<0的距離為(

)A.3 B.4 C.5 D.6【答案】B【解析】【分析】過點(diǎn)SKIPIF1<0作SKIPIF1<0軸的垂線,垂足為SKIPIF1<0,進(jìn)而根據(jù)SKIPIF1<0得SKIPIF1<0,再結(jié)合拋物線定義即可得答案.【詳解】解:如圖,過點(diǎn)SKIPIF1<0作SKIPIF1<0軸的垂線,垂足為SKIPIF1<0,由題知SKIPIF1<0,即SKIPIF1<0因?yàn)镾KIPIF1<0,所以SKIPIF1<0所以SKIPIF1<0,所以點(diǎn)SKIPIF1<0到準(zhǔn)線SKIPIF1<0的距離為SKIPIF1<0.故選:B7.(2022·天津南開·二模)設(shè)拋物線SKIPIF1<0的焦點(diǎn)到雙曲線SKIPIF1<0的一條漸近線的距離為SKIPIF1<0,到雙曲線左頂點(diǎn)的距離為SKIPIF1<0,則該雙曲線的離心率是(

)A.SKIPIF1<0 B.SKIPIF1<0 C.2 D.SKIPIF1<0【答案】C【解析】【分析】先得到拋物線SKIPIF1<0的焦點(diǎn)坐標(biāo),然后根據(jù)題意,利用點(diǎn)到直線的距離和兩點(diǎn)間的距離求解.【詳解】解:拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0,設(shè)雙曲線SKIPIF1<0的一條漸近線方程為SKIPIF1<0,由題意得SKIPIF1<0,解得SKIPIF1<0,雙曲線左頂點(diǎn)為SKIPIF1<0,由題意得SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,所以該雙曲線的離心率是SKIPIF1<0,故選:C8.(2022·陜西西安·三模(理))已知拋物線SKIPIF1<0上一點(diǎn)SKIPIF1<0,SKIPIF1<0為其焦點(diǎn),直線SKIPIF1<0交拋物線的準(zhǔn)線于點(diǎn)SKIPIF1<0.且線段SKIPIF1<0的中點(diǎn)為SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】設(shè)點(diǎn)SKIPIF1<0,利用中點(diǎn)坐標(biāo)公式求出SKIPIF1<0的值,可得出拋物線的方程,再將點(diǎn)SKIPIF1<0的坐標(biāo)代入拋物線的方程,可求得SKIPIF1<0的值.【詳解】拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0,準(zhǔn)線方程為SKIPIF1<0,設(shè)點(diǎn)SKIPIF1<0,由題意可得SKIPIF1<0,解得SKIPIF1<0,所以拋物線的方程為SKIPIF1<0,所以,SKIPIF1<0,解得SKIPIF1<0.故選:D.二、多選題1.(2022·湖南常德·一模)已知拋物線SKIPIF1<0的焦點(diǎn)SKIPIF1<0到準(zhǔn)線SKIPIF1<0的距離為2,則(

)A.焦點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0B.過點(diǎn)SKIPIF1<0恰有2條直線與拋物線SKIPIF1<0有且只有一個(gè)公共點(diǎn)C.直線SKIPIF1<0與拋物線SKIPIF1<0相交所得弦長(zhǎng)為8D.拋物線SKIPIF1<0與圓SKIPIF1<0交于SKIPIF1<0兩點(diǎn),則SKIPIF1<0【答案】ACD【解析】【分析】先求出拋物線方程,對(duì)選項(xiàng)逐一判斷即可.【詳解】由題可知拋物線方程為SKIPIF1<0對(duì)于A,焦點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0,故A正確對(duì)于B,過點(diǎn)SKIPIF1<0有拋物線的2條切線,還有SKIPIF1<0,共3條直線與拋物線有且只有一個(gè)交點(diǎn),故B錯(cuò)誤對(duì)于C,SKIPIF1<0,弦長(zhǎng)為SKIPIF1<0,故C正確對(duì)于D,SKIPIF1<0,解得SKIPIF1<0(SKIPIF1<0舍去),交點(diǎn)為SKIPIF1<0,有SKIPIF1<0,故D正確故選:ACD2.(2022·湖南·雅禮中學(xué)二模)在平面直角坐標(biāo)系SKIPIF1<0中,點(diǎn)SKIPIF1<0在拋物線SKIPIF1<0上,拋物線的焦點(diǎn)為SKIPIF1<0,延長(zhǎng)SKIPIF1<0與拋物線相交于點(diǎn)SKIPIF1<0,則下列結(jié)論正確的是(

)A.拋物線的準(zhǔn)線方程為SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0的面積為SKIPIF1<0 D.SKIPIF1<0【答案】AD【解析】根據(jù)條件求出SKIPIF1<0,再聯(lián)立直線與拋物線求出SKIPIF1<0,進(jìn)而求出結(jié)論.【詳解】解:SKIPIF1<0點(diǎn)SKIPIF1<0在拋物線SKIPIF1<0上,SKIPIF1<0,SKIPIF1<0,焦點(diǎn)為SKIPIF1<0,準(zhǔn)線為SKIPIF1<0,SKIPIF1<0對(duì),因?yàn)镾KIPIF1<0,故SKIPIF1<0,故直線SKIPIF1<0為:SKIPIF1<0,聯(lián)立SKIPIF1<0SKIPIF1<0SKIPIF1<0或SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0錯(cuò),SKIPIF1<0,SKIPIF1<0對(duì),SKIPIF1<0的面積為SKIPIF1<0.故SKIPIF1<0錯(cuò),故選:SKIPIF1<0.三、解答題1.(2022·江西萍鄉(xiāng)·三模(理))已知拋物線SKIPIF1<0的頂點(diǎn)在坐標(biāo)原點(diǎn),焦點(diǎn)與圓SKIPIF1<0的圓心重合,SKIPIF1<0為SKIPIF1<0上一動(dòng)點(diǎn),點(diǎn)SKIPIF1<0.若SKIPIF1<0的最小值為SKIPIF1<0.(1)求拋物線SKIPIF1<0的標(biāo)準(zhǔn)方程;(2)過焦點(diǎn)的直線SKIPIF1<0與拋物線SKIPIF1<0和圓SKIPIF1<0從左向右依次交于SKIPIF1<0四點(diǎn),且滿足SKIPIF1<0,求直線SKIPIF1<0的方程.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)由圓的方程可確定焦點(diǎn)坐標(biāo),設(shè)拋物線為SKIPIF1<0;過SKIPIF1<0作拋物線準(zhǔn)線的垂線,由拋物

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