新高考數(shù)學(xué)一輪復(fù)習(xí) 講與練第27講 橢圓（原卷版）

第27講橢圓學(xué)校____________姓名____________班級(jí)____________知識(shí)梳理1.橢圓的定義如果F1，F(xiàn)2是平面內(nèi)的兩個(gè)定點(diǎn)，a是一個(gè)常數(shù)，且2a＞|F1F2|，則平面內(nèi)滿(mǎn)足|PF1|＋|PF2|＝2a的動(dòng)點(diǎn)P的軌跡稱(chēng)為橢圓，其中兩個(gè)定點(diǎn)F1，F(xiàn)2稱(chēng)為橢圓的焦點(diǎn)，兩個(gè)焦點(diǎn)之間的距離|F1F2|稱(chēng)為橢圓的焦距.其數(shù)學(xué)表達(dá)式：集合M＝{P||PF1|＋|PF2|＝2a}，|F1F2|＝2c，其中a＞0，c＞0，且a，c為常數(shù)：(1)若a＞c，則點(diǎn)P的軌跡為橢圓；(2)若a＝c，則點(diǎn)P的軌跡為線(xiàn)段；(3)若a＜c，則點(diǎn)P的軌跡不存在.2.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(chēng)性對(duì)稱(chēng)軸：坐標(biāo)軸；對(duì)稱(chēng)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考點(diǎn)和典型例題1、橢圓的定義及應(yīng)用【典例1-1】已知SKIPIF1<0，SKIPIF1<0是兩個(gè)定點(diǎn)，且SKIPIF1<0（SKIPIF1<0是正常數(shù)），動(dòng)點(diǎn)SKIPIF1<0滿(mǎn)足SKIPIF1<0，則動(dòng)點(diǎn)SKIPIF1<0的軌跡是（

）A．橢圓 B．線(xiàn)段 C．橢圓或線(xiàn)段 D．直線(xiàn)【答案】C【詳解】解：因?yàn)镾KIPIF1<0（當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)動(dòng)點(diǎn)SKIPIF1<0的軌跡是橢圓；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)動(dòng)點(diǎn)SKIPIF1<0的軌跡是線(xiàn)段SKIPIF1<0．故選：C．【典例1-2】已知橢圓SKIPIF1<0的兩個(gè)焦點(diǎn)為SKIPIF1<0，SKIPIF1<0，過(guò)SKIPIF1<0的直線(xiàn)交橢圓于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，若SKIPIF1<0的周長(zhǎng)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】由SKIPIF1<0.因?yàn)镾KIPIF1<0，SKIPIF1<0是橢圓的上的點(diǎn)，SKIPIF1<0、SKIPIF1<0是橢圓的焦點(diǎn)，所以SKIPIF1<0，因此SKIPIF1<0的周長(zhǎng)為SKIPIF1<0，故選：D【典例1-3】已知橢圓SKIPIF1<0的兩個(gè)焦點(diǎn)分別為SKIPIF1<0是橢圓上一點(diǎn)，SKIPIF1<0，且離心率為SKIPIF1<0，則橢圓C的標(biāo)準(zhǔn)方程為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】根據(jù)橢圓定義可得SKIPIF1<0，所以SKIPIF1<0，由離心率SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0，所以橢圓C的標(biāo)準(zhǔn)方程為SKIPIF1<0.故選：B【典例1-4】已知橢圓SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0，點(diǎn)SKIPIF1<0在橢圓上，若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】解：由題意，橢圓方程SKIPIF1<0，可得SKIPIF1<0，所以焦點(diǎn)SKIPIF1<0，又由橢圓的定義，可得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，在SKIPIF1<0中，由余弦定理可得SKIPIF1<0,所以SKIPIF1<0，解得SKIPIF1<0，又由SKIPIF1<0，所以SKIPIF1<0.故選:C．【典例1-5】已知點(diǎn)SKIPIF1<0在橢圓SKIPIF1<0上，SKIPIF1<0與SKIPIF1<0分別為左、右焦點(diǎn)，若SKIPIF1<0，則SKIPIF1<0的面積為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】由SKIPIF1<0，，又SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0.故選：A.2、橢圓的簡(jiǎn)單幾何性質(zhì)【典例2-1】橢圓SKIPIF1<0的左右焦點(diǎn)分別為SKIPIF1<0，右頂點(diǎn)為SKIPIF1<0，點(diǎn)SKIPIF1<0在橢圓上，滿(mǎn)足SKIPIF1<0，則橢圓的離心率為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】由SKIPIF1<0，SKIPIF1<0得SKIPIF1<0,故SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，在△SKIPIF1<0中，由余弦定理可得：SKIPIF1<0，SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0解得SKIPIF1<0或SKIPIF1<0（舍），故選：B.【典例2-2】橢圓SKIPIF1<0：SKIPIF1<0與雙曲線(xiàn)SKIPIF1<0：SKIPIF1<0的離心率之積為1，則雙曲線(xiàn)SKIPIF1<0的兩條漸近線(xiàn)的傾斜角分別為（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0 C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】D【詳解】因?yàn)闄E圓SKIPIF1<0：SKIPIF1<0與雙曲線(xiàn)SKIPIF1<0：SKIPIF1<0的離心率之積為1，所以有SKIPIF1<0，因此雙曲線(xiàn)SKIPIF1<0的兩條漸近線(xiàn)方程為：SKIPIF1<0，所以雙曲線(xiàn)SKIPIF1<0的兩條漸近線(xiàn)的傾斜角分別為SKIPIF1<0，SKIPIF1<0，故選：D【典例2-3】已知點(diǎn)A、B為橢圓SKIPIF1<0的長(zhǎng)軸頂點(diǎn)，P為橢圓上一點(diǎn)，若直線(xiàn)PA，PB的斜率之積的范圍為SKIPIF1<0，則橢圓SKIPIF1<0的離心率的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】由題得：SKIPIF1<0，所以SKIPIF1<0故選：A．【典例2-4】已知雙曲線(xiàn)SKIPIF1<0的左、右頂點(diǎn)為SKIPIF1<0，SKIPIF1<0，焦點(diǎn)在y軸上的橢圓以SKIPIF1<0，SKIPIF1<0為頂點(diǎn)，且離心率為SKIPIF1<0，過(guò)SKIPIF1<0作斜率為SKIPIF1<0的直線(xiàn)SKIPIF1<0交雙曲線(xiàn)于另一點(diǎn)SKIPIF1<0，交橢圓于另一點(diǎn)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】設(shè)所求橢圓的標(biāo)準(zhǔn)方程為SKIPIF1<0，半焦距為SKIPIF1<0，雙曲線(xiàn)SKIPIF1<0的左頂點(diǎn)為SKIPIF1<0，右頂點(diǎn)為SKIPIF1<0，由于橢圓SKIPIF1<0以SKIPIF1<0，SKIPIF1<0為頂點(diǎn)，則SKIPIF1<0，該橢圓的離心率為SKIPIF1<0，所以，SKIPIF1<0，解得SKIPIF1<0，所以，橢圓的方程為SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0，由于SKIPIF1<0，則點(diǎn)SKIPIF1<0，由于點(diǎn)SKIPIF1<0在橢圓上，點(diǎn)SKIPIF1<0在雙曲線(xiàn)上，所以，SKIPIF1<0，聯(lián)立得：SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0，所以SKIPIF1<0，此時(shí)點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0重合，不滿(mǎn)足題意舍去；當(dāng)SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：B.【典例2-5】已知橢圓SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0、SKIPIF1<0，第一象限內(nèi)的點(diǎn)SKIPIF1<0在橢圓上，且滿(mǎn)足SKIPIF1<0，點(diǎn)SKIPIF1<0在線(xiàn)段SKIPIF1<0、SKIPIF1<0上，設(shè)SKIPIF1<0，將SKIPIF1<0沿SKIPIF1<0翻折，使得平面SKIPIF1<0與平面SKIPIF1<0垂直，要使翻折后SKIPIF1<0的長(zhǎng)度最小，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】在橢圓SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，且點(diǎn)SKIPIF1<0為第一象限內(nèi)的點(diǎn)，則SKIPIF1<0，可得SKIPIF1<0，翻折前，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0，垂足為點(diǎn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0，垂足為點(diǎn)SKIPIF1<0，設(shè)SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以，SKIPIF1<0，翻折后，如下圖所示：因?yàn)槠矫鍿KIPIF1<0平面SKIPIF1<0，平面SKIPIF1<0平面SKIPIF1<0，SKIPIF1<0平面SKIPIF1<0，SKIPIF1<0，SKIPIF1<0平面SKIPIF1<0，SKIPIF1<0平面SKIPIF1<0，SKIPIF1<0，又因?yàn)镾KIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，即當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最小值SKIPIF1<0，則在翻折前，在SKIPIF1<0中，SKIPIF1<0為SKIPIF1<0的角平分線(xiàn)，所以，SKIPIF1<0，即SKIPIF1<0.故選：A.3、橢圓的綜合應(yīng)用【典例3-1】（多選）已知橢圓SKIPIF1<0的左，右焦點(diǎn)分別為SKIPIF1<0，橢圓的上頂點(diǎn)和右頂點(diǎn)分別為A，B．若P，Q兩點(diǎn)都在橢圓C上，且P，Q關(guān)于坐標(biāo)原點(diǎn)對(duì)稱(chēng)，則（

）A．|PQ|的最大值為SKIPIF1<0B．SKIPIF1<0為定值C．橢圓上不存在點(diǎn)M，使得SKIPIF1<0D．若點(diǎn)P在第一象限，則四邊形APBQ面積的最大值為SKIPIF1<0【答案】BD【詳解】如圖所示：A.|PQ|的最大值為長(zhǎng)軸長(zhǎng)2SKIPIF1<0，故錯(cuò)誤；B.易知SKIPIF1<0是平行四邊形，則SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，故正確；C.因?yàn)镾KIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，故橢圓上存在點(diǎn)M，使得SKIPIF1<0，故錯(cuò)誤；D.直線(xiàn)AB所在直線(xiàn)方程為：SKIPIF1<0，即SKIPIF1<0，設(shè)SKIPIF1<0，則點(diǎn)P到直線(xiàn)AB的距離為SKIPIF1<0，其最大值為SKIPIF1<0，同理點(diǎn)Q到直線(xiàn)AB的最大值為SKIPIF1<0，所以四邊形APBQ面積的最大值為SKIPIF1<0，故正確.故選：BD【典例3-2】（多選）過(guò)橢圓SKIPIF1<0的中心任作一直線(xiàn)交橢圓于P，Q兩點(diǎn)，SKIPIF1<0，SKIPIF1<0是橢圓的左、右焦點(diǎn)，A，B是橢圓的左、右頂點(diǎn)，則下列說(shuō)法正確的是（

）A．SKIPIF1<0周長(zhǎng)的最小值為18B．四邊形SKIPIF1<0可能為矩形C．若直線(xiàn)PA斜率的取值范圍是SKIPIF1<0，則直線(xiàn)PB斜率的取值范圍是SKIPIF1<0D．SKIPIF1<0的最小值為-1【答案】AC【詳解】A：根據(jù)橢圓的對(duì)稱(chēng)性，SKIPIF1<0，當(dāng)PQ為橢圓的短軸時(shí)，SKIPIF1<0有最小值8，所以SKIPIF1<0周長(zhǎng)的最小值為18，正確；B：若四邊形SKIPIF1<0為矩形，則點(diǎn)P，Q必在以SKIPIF1<0為直徑的圓上，但此圓與橢圓SKIPIF1<0無(wú)交點(diǎn)，錯(cuò)誤；C：設(shè)SKIPIF1<0，則SKIPIF1<0，因?yàn)橹本€(xiàn)PA斜率的范圍是SKIPIF1<0，所以直線(xiàn)PB斜率的范圍是SKIPIF1<0，正確；D：設(shè)SKIPIF1<0，則SKIPIF1<0SKIPIF1<0SKIPIF1<0．因?yàn)镾KIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0最小值為SKIPIF1<0，錯(cuò)誤.故選：AC．【典例3-3】（多選）已知橢圓SKIPIF1<0的左，右焦點(diǎn)分別為SKIPIF1<0，A，B兩點(diǎn)都在C上，且A，B關(guān)于坐標(biāo)原點(diǎn)對(duì)稱(chēng)，則（

）A．SKIPIF1<0的最大值為SKIPIF1<0 B．SKIPIF1<0為定值C．C的焦距是短軸長(zhǎng)的2倍 D．存在點(diǎn)A，使得SKIPIF1<0【答案】ABD【詳解】解：由題意，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以A正確，C錯(cuò)誤；由橢圓的對(duì)稱(chēng)性知，SKIPIF1<0，所以B正確；當(dāng)A在y軸上時(shí)，SKIPIF1<0，則SKIPIF1<0為鈍角，所以存在點(diǎn)A，使得SKIPIF1<0，所以D正確.故選：ABD.【典例3-4】已知橢圓SKIPIF1<0的兩焦點(diǎn)分別為SKIPIF1<0和SKIPIF1<0，短軸的一個(gè)端點(diǎn)為SKIPIF1<0．(1)求橢圓C的標(biāo)準(zhǔn)方程和離心率；(2)橢圓C上是否存在一點(diǎn)P，使得SKIPIF1<0?若存在，求SKIPIF1<0的面積；若不存在，請(qǐng)說(shuō)明理由．【解析】(1)由焦點(diǎn)坐標(biāo)知SKIPIF1<0，由短軸端點(diǎn)SKIPIF1<0知SKIPIF1<0，所以SKIPIF1<0，故所求橢圓標(biāo)準(zhǔn)方程為SKIPIF1<0.(2)假設(shè)橢圓C上存在一點(diǎn)SKIPIF1<0，使得SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，聯(lián)立SKIPIF1<0，得SKIPIF1<0，此方程無(wú)解.故橢圓上不存在點(diǎn)P，