新高考數(shù)學(xué)二輪復(fù)習(xí)圓錐曲線重難點(diǎn)提升專題3 圓錐曲線中的長(zhǎng)度問(wèn)題（原卷版）

專題3圓錐曲線中的長(zhǎng)度問(wèn)題一、考情分析圓錐曲線中的長(zhǎng)度問(wèn)題是直線與圓錐曲線中最基本的問(wèn)題,一般出現(xiàn)在解答題第2問(wèn),常見(jiàn)的有焦半徑、弦長(zhǎng)、兩點(diǎn)間距離、點(diǎn)到直線距離、三角形周長(zhǎng)等,求解方法可以用兩點(diǎn)間距離公式、弦長(zhǎng)公式、點(diǎn)到直線距離公式、函數(shù)求最值等.二、解題秘籍(一)利用兩點(diǎn)間距離公式求線段長(zhǎng)度若直線與圓錐曲線的交點(diǎn)坐標(biāo)已知或可求,可直線利用兩點(diǎn)間距離公式求線段長(zhǎng)度.【例1】（2022屆山西省呂梁市高三上學(xué)期12月月考）在平面直角坐標(biāo)系SKIPIF1<0中,已知橢圓SKIPIF1<0的右準(zhǔn)線為SKIPIF1<0（定義：橢圓C的右準(zhǔn)線方程為SKIPIF1<0,其中SKIPIF1<0）.點(diǎn)P是右準(zhǔn)線上的動(dòng)點(diǎn),過(guò)點(diǎn)P作橢圓C的兩條切線,分別與y軸交于M,N兩點(diǎn).當(dāng)P在x軸上時(shí),SKIPIF1<0.（1）求橢圓C的方程；（2）求SKIPIF1<0的最小值.【解析】（1）由題意可知,當(dāng)P點(diǎn)坐標(biāo)為SKIPIF1<0時(shí),SKIPIF1<0,不妨設(shè)點(diǎn)M在點(diǎn)N上方,則SKIPIF1<0,所以直線SKIPIF1<0與橢圓C相切,將直線SKIPIF1<0與橢圓方程聯(lián)立,SKIPIF1<0消去y,整理得SKIPIF1<0,則SKIPIF1<0,整理得SKIPIF1<0,又SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0（舍去）,所以SKIPIF1<0,即橢圓C的方程為SKIPIF1<0；（2）設(shè)SKIPIF1<0,切線方程為SKIPIF1<0,將切線方程與橢圓聯(lián)立,SKIPIF1<0消去y,整理得SKIPIF1<0,則SKIPIF1<0,整理得SKIPIF1<0,設(shè)切線SKIPIF1<0斜率為SKIPIF1<0,直線SKIPIF1<0斜率為SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,將SKIPIF1<0,SKIPIF1<0代入上式,整理得SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),上述等號(hào)成立,即SKIPIF1<0的最小值為4.(二)利用SKIPIF1<0求距離設(shè)斜率為k(k≠0)的直線l與圓錐曲線C相交于A(x1,y1),B(x2,y2)兩點(diǎn),則|AB|＝eq\r(1＋k2)|x2－x1|.其中求|x2－x1|通常使用根與系數(shù)的關(guān)系,即作如下變形：|x2－x1|＝SKIPIF1<0,【例2】(2022屆陜西省安康市高三下學(xué)期聯(lián)考)已知橢圓SKIPIF1<0長(zhǎng)軸的頂點(diǎn)與雙曲線SKIPIF1<0實(shí)軸的頂點(diǎn)相同,且SKIPIF1<0的右焦點(diǎn)SKIPIF1<0到SKIPIF1<0的漸近線的距離為SKIPIF1<0．(1)求SKIPIF1<0與SKIPIF1<0的方程；(2)若直線SKIPIF1<0的傾斜角是直線SKIPIF1<0的傾斜角的SKIPIF1<0倍,且SKIPIF1<0經(jīng)過(guò)點(diǎn)SKIPIF1<0,SKIPIF1<0與SKIPIF1<0交于SKIPIF1<0、SKIPIF1<0兩點(diǎn),與SKIPIF1<0交于SKIPIF1<0、SKIPIF1<0兩點(diǎn),求SKIPIF1<0．【解析】（1）由題意可得SKIPIF1<0,則SKIPIF1<0．因?yàn)镾KIPIF1<0的漸近線方程為SKIPIF1<0,即SKIPIF1<0,橢圓SKIPIF1<0的右焦點(diǎn)為SKIPIF1<0,由題意可得SKIPIF1<0,SKIPIF1<0,解得SKIPIF1<0,故橢圓SKIPIF1<0的方程為SKIPIF1<0,雙曲線SKIPIF1<0的方程為SKIPIF1<0．（2）設(shè)直線SKIPIF1<0的傾斜角為SKIPIF1<0,所以,直線SKIPIF1<0的斜率為SKIPIF1<0,所以直線SKIPIF1<0的方程為SKIPIF1<0,聯(lián)立SKIPIF1<0得SKIPIF1<0,則SKIPIF1<0,設(shè)SKIPIF1<0、SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,聯(lián)立SKIPIF1<0可得SKIPIF1<0,SKIPIF1<0,設(shè)點(diǎn)SKIPIF1<0、SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,所以,SKIPIF1<0,故SKIPIF1<0．(三)利用SKIPIF1<0求距離設(shè)斜率為k(k≠0)的直線l與圓錐曲線C相交于A(x1,y1),B(x2,y2)兩點(diǎn),則|AB|＝eq\r(1＋\f(1,k2))|y2－y1|.當(dāng)消去x整理方程為關(guān)于y的一元二次方程常用此結(jié)論.其中求|y2－y1|時(shí)通常使用根與系數(shù)的關(guān)系,即作如下變形：|y2－y1|＝SKIPIF1<0.【例3】（2023屆重慶市巴蜀中學(xué)校高三上學(xué)期月考）已知橢圓SKIPIF1<0的離心率SKIPIF1<0；上頂點(diǎn)為A,右頂點(diǎn)為SKIPIF1<0,直線SKIPIF1<0與圓SKIPIF1<0相切.(1)求橢圓SKIPIF1<0的標(biāo)準(zhǔn)方程；(2)設(shè)與圓SKIPIF1<0相切的直線SKIPIF1<0與橢圓相交于SKIPIF1<0兩點(diǎn),SKIPIF1<0為弦SKIPIF1<0的中點(diǎn),SKIPIF1<0為坐標(biāo)原點(diǎn).求SKIPIF1<0的取值范圍.【解析】（1）由SKIPIF1<0知SKIPIF1<0,原點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0,故SKIPIF1<0,故橢圓SKIPIF1<0的標(biāo)準(zhǔn)方程為SKIPIF1<0.（2）SKIPIF1<0時(shí)：SKIPIF1<0,或SKIPIF1<0,故SKIPIF1<0；直線SKIPIF1<0斜率不存在時(shí),SKIPIF1<0,或SKIPIF1<0．故SKIPIF1<0；直線SKIPIF1<0斜率存在且不為0時(shí)：設(shè)直線SKIPIF1<0的方程為SKIPIF1<0（SKIPIF1<0）,由直線SKIPIF1<0與圓SKIPIF1<0相切,所以SKIPIF1<0,即SKIPIF1<0,聯(lián)立SKIPIF1<0得SKIPIF1<0,設(shè)SKIPIF1<0,由韋達(dá)定理：SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0中點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0,故SKIPIF1<0SKIPIF1<0SKIPIF1<0,故SKIPIF1<0SKIPIF1<0,SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,SKIPIF1<0時(shí)等號(hào)成立,SKIPIF1<0綜上：SKIPIF1<0的取值范圍是SKIPIF1<0.(四)利用點(diǎn)到直線距離公式求垂線段的長(zhǎng)1.若已知定點(diǎn)P,點(diǎn)Q在動(dòng)直線上,求SKIPIF1<0最小值,常利用點(diǎn)到直線距離公式；2.若點(diǎn)P在定直線上,點(diǎn)Q為曲線上,求SKIPIF1<0最小值,有時(shí)可轉(zhuǎn)換為與定直線平行的切線的切點(diǎn)到定直線的距離.【例4】（2023屆上海市華東師范大學(xué)第二附屬中學(xué)高三上學(xué)期考試）設(shè)有橢圓方程SKIPIF1<0,直線SKIPIF1<0,SKIPIF1<0下端點(diǎn)為SKIPIF1<0,左?右焦點(diǎn)分別為SKIPIF1<0在SKIPIF1<0上.(1)若SKIPIF1<0中點(diǎn)在SKIPIF1<0軸上,求點(diǎn)SKIPIF1<0的坐標(biāo)；(2)直線SKIPIF1<0與SKIPIF1<0軸交于SKIPIF1<0,直線SKIPIF1<0經(jīng)過(guò)右焦點(diǎn)SKIPIF1<0,且SKIPIF1<0,求SKIPIF1<0；(3)在橢圓SKIPIF1<0上存在一點(diǎn)SKIPIF1<0到SKIPIF1<0距離為SKIPIF1<0,使SKIPIF1<0,當(dāng)SKIPIF1<0變化時(shí),求SKIPIF1<0的最小值.【解析】（1）因?yàn)樽蠼裹c(diǎn)SKIPIF1<0,所以SKIPIF1<0,由題知SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,又因?yàn)镾KIPIF1<0中點(diǎn)在SKIPIF1<0軸上,所以點(diǎn)SKIPIF1<0的縱坐標(biāo)為1,代入SKIPIF1<0中的SKIPIF1<0,所以點(diǎn)SKIPIF1<0坐標(biāo)為SKIPIF1<0.（2）如圖,設(shè)直線SKIPIF1<0與SKIPIF1<0軸交點(diǎn)為SKIPIF1<0,因?yàn)橹本€SKIPIF1<0為SKIPIF1<0,所以直線SKIPIF1<0的傾斜角為SKIPIF1<0,SKIPIF1<0①,由題意知,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以在SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,整理可得SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,又因?yàn)镾KIPIF1<0,所以SKIPIF1<0,SKIPIF1<0舍去,SKIPIF1<0.（3）設(shè)直線SKIPIF1<0平移后與橢圓相切的直線SKIPIF1<0方程為SKIPIF1<0,聯(lián)立SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,因?yàn)闄E圓上存在點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0所以SKIPIF1<0①,同時(shí)SKIPIF1<0,又因?yàn)镾KIPIF1<0,所以①式右側(cè)肯定成立,左側(cè)可以整理為SKIPIF1<0,解得SKIPIF1<0,因?yàn)镾KIPIF1<0,所以當(dāng)SKIPIF1<0取得最小值SKIPIF1<0時(shí),SKIPIF1<0有最大值,最大值為SKIPIF1<0.(五)利用函數(shù)思想求距離最值求圓錐曲線上的動(dòng)點(diǎn)到一定點(diǎn)距離的最值,有時(shí)可設(shè)出動(dòng)點(diǎn)坐標(biāo),利用距離公式把問(wèn)題轉(zhuǎn)化為函數(shù)求最值.【例5】已知橢圓SKIPIF1<0的長(zhǎng)軸長(zhǎng)為SKIPIF1<0,點(diǎn)SKIPIF1<0在SKIPIF1<0上.（1）求SKIPIF1<0的方程；（2）設(shè)SKIPIF1<0的上頂點(diǎn)為A,右頂點(diǎn)為B,直線SKIPIF1<0與SKIPIF1<0平行,且與SKIPIF1<0交于SKIPIF1<0,SKIPIF1<0兩點(diǎn),SKIPIF1<0,點(diǎn)SKIPIF1<0為SKIPIF1<0的右焦點(diǎn),求SKIPIF1<0的最小值.【解析】（1）因?yàn)镾KIPIF1<0的長(zhǎng)軸長(zhǎng)為SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0.又點(diǎn)SKIPIF1<0在SKIPIF1<0上,所以SKIPIF1<0,代入SKIPIF1<0,解得SKIPIF1<0,故SKIPIF1<0的方程為SKIPIF1<0.（2）由（1）可知,A,B的坐標(biāo)分別為SKIPIF1<0,SKIPIF1<0,直線SKIPIF1<0的方程為SKIPIF1<0,設(shè)SKIPIF1<0,聯(lián)立SKIPIF1<0得SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0,設(shè)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,因?yàn)镾KIPIF1<0,所以D為MN的中點(diǎn),則SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0的坐標(biāo)為SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0,因?yàn)镾KIPIF1<0,所以當(dāng)SKIPIF1<0時(shí),SKIPIF1<0取得最小值,且最小值為SKIPIF1<0.(六)利用圓錐曲線定義求長(zhǎng)度與圓錐曲線焦點(diǎn)弦或焦半徑有關(guān)的長(zhǎng)度計(jì)算可利用圓錐曲線定義求解.【例6】（2022屆湖南省長(zhǎng)沙市寧鄉(xiāng)市高三下學(xué)期5月模擬）已知拋物線SKIPIF1<0SKIPIF1<0的焦點(diǎn)與橢圓SKIPIF1<0SKIPIF1<0的右焦點(diǎn)SKIPIF1<0重合,橢圓SKIPIF1<0的長(zhǎng)軸長(zhǎng)為SKIPIF1<0.(1)求橢圓SKIPIF1<0的方程；(2)過(guò)點(diǎn)SKIPIF1<0且斜率為SKIPIF1<0的直線SKIPIF1<0交橢圓SKIPIF1<0于SKIPIF1<0兩點(diǎn),交拋物線SKIPIF1<0于SKIPIF1<0兩點(diǎn),請(qǐng)問(wèn)是否存在實(shí)常數(shù)SKIPIF1<0,使SKIPIF1<0為定值？若存在,求出SKIPIF1<0的值；若不存在,說(shuō)明理由.【解析】（1）因?yàn)閽佄锞€SKIPIF1<0SKIPIF1<0的焦點(diǎn)為SKIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0,則SKIPIF1<0,故橢圓SKIPIF1<0的方程為：SKIPIF1<0；（2）設(shè)SKIPIF1<0?SKIPIF1<0?SKIPIF1<0?SKIPIF1<0,設(shè)直線SKIPIF1<0的方程為SKIPIF1<0,與橢圓SKIPIF1<0的方程聯(lián)立SKIPIF1<0,得SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,設(shè)直線SKIPIF1<0的方程SKIPIF1<0,與拋物線G的方程聯(lián)立SKIPIF1<0,得SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,要使SKIPIF1<0為常數(shù),則SKIPIF1<0,解得SKIPIF1<0,故存在SKIPIF1<0,使得SKIPIF1<0為定值SKIPIF1<0．【例7】（2023屆江蘇省南京市高三上學(xué)期測(cè)試）已知點(diǎn)B是圓C:SKIPIF1<0上的任意一點(diǎn),點(diǎn)F（SKIPIF1<0,0）,線段BF的垂直平分線交BC于點(diǎn)P.(1)求動(dòng)點(diǎn)Р的軌跡E的方程;(2)設(shè)曲線E與x軸的兩個(gè)交點(diǎn)分別為A1,A2,Q為直線x=4上的動(dòng)點(diǎn),且Q不在x軸上,QA1與E的另一個(gè)交點(diǎn)為M,QA2與E的另一個(gè)交點(diǎn)為N,證明:△FMN的周長(zhǎng)為定值.【解析】（1）因?yàn)辄c(diǎn)P在BF垂直平分線上,所以有SKIPIF1<0,所以：SKIPIF1<0,即PF+PC為定值4SKIPIF1<0,所以軌跡E為橢圓,且SKIPIF1<0,所以SKIPIF1<0,所以軌跡E的方程為：SKIPIF1<0.（2）由題知：SKIPIF1<0,設(shè)SKIPIF1<0則SKIPIF1<0,SKIPIF1<0,所以QA1方程為：SKIPIF1<0,QA2方程為：SKIPIF1<0,聯(lián)立方程：SKIPIF1<0,可以得出M：SKIPIF1<0同理可以計(jì)算出點(diǎn)N坐標(biāo)：SKIPIF1<0,當(dāng)SKIPIF1<0存在,即SKIPIF1<0,即SKIPIF1<0時(shí),SKIPIF1<0所以直線MN的方程為：SKIPIF1<0即：SKIPIF1<0,所以直線過(guò)定點(diǎn)SKIPIF1<0,即過(guò)橢圓的右焦點(diǎn)SKIPIF1<0,所以△FMN的周長(zhǎng)為4a=8.當(dāng)SKIPIF1<0不存在,即SKIPIF1<0,即SKIPIF1<0時(shí),可以計(jì)算出SKIPIF1<0,周長(zhǎng)也等于8.所以△FMN的周長(zhǎng)為定值8.三、跟蹤檢測(cè)1．（2023屆北京市高三上學(xué)期入學(xué)定位考試）已知橢圓C：SKIPIF1<0（其中SKIPIF1<0）的離心率為SKIPIF1<0,左右焦點(diǎn)分別為SKIPIF1<0,SKIPIF1<0.(1)求橢圓C的方程；(2)過(guò)點(diǎn)SKIPIF1<0作斜率為k的直線與橢圓C交于不同的A,B兩點(diǎn),過(guò)原點(diǎn)作AB的垂線,垂足為D.若點(diǎn)D恰好是SKIPIF1<0與A的中點(diǎn),求線段AB的長(zhǎng)度.2．（2023屆福建省部分名校高三上學(xué)期9月聯(lián)考）已知兩點(diǎn)SKIPIF1<0,SKIPIF1<0,動(dòng)點(diǎn)SKIPIF1<0在SKIPIF1<0軸的投影為SKIPIF1<0,且SKIPIF1<0,記動(dòng)點(diǎn)SKIPIF1<0的軌跡為曲線SKIPIF1<0.(1)求SKIPIF1<0的方程.(2)過(guò)點(diǎn)SKIPIF1<0的直線與曲線SKIPIF1<0在SKIPIF1<0軸右側(cè)相交于SKIPIF1<0,SKIPIF1<0兩點(diǎn),線段SKIPIF1<0的垂直平分線與SKIPIF1<0軸相交于點(diǎn)SKIPIF1<0,試問(wèn)SKIPIF1<0是否為定值？若是,求出該定值；若不是,請(qǐng)說(shuō)明理由.3．（2023屆四川省巴中市高三上學(xué)期考試）已知橢圓SKIPIF1<0：SKIPIF1<0的左、右頂點(diǎn)分別為SKIPIF1<0、SKIPIF1<0,點(diǎn)SKIPIF1<0在橢圓SKIPIF1<0上,且直線SKIPIF1<0的斜率與直線SKIPIF1<0的斜率之積為SKIPIF1<0．(1)求橢圓SKIPIF1<0的方程；(2)若圓SKIPIF1<0的切線SKIPIF1<0與橢圓SKIPIF1<0交于SKIPIF1<0、SKIPIF1<0兩點(diǎn),求SKIPIF1<0的最大值及此時(shí)直線SKIPIF1<0的斜率．4．（2023屆安徽省部分校高三上學(xué)期摸底考）已知SKIPIF1<0為坐標(biāo)原點(diǎn),橢圓SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0,記線段SKIPIF1<0的中點(diǎn)為SKIPIF1<0．(1)若直線SKIPIF1<0的斜率為3,求直線SKIPIF1<0的斜率;(2)若四邊形SKIPIF1<0為平行四邊形,求SKIPIF1<0的取值范圍．5.（2023屆遼寧省朝陽(yáng)市高三上學(xué)期9月月考）已知雙曲線SKIPIF1<0的離心率為SKIPIF1<0,點(diǎn)SKIPIF1<0在雙曲線SKIPIF1<0上.(1)求雙曲線SKIPIF1<0的方程；(2)點(diǎn)SKIPIF1<0,SKIPIF1<0在雙曲線SKIPIF1<0上,直線SKIPIF1<0,SKIPIF1<0與SKIPIF1<0軸分別相交于SKIPIF1<0兩點(diǎn),點(diǎn)SKIPIF1<0在直線SKIPIF1<0上,若坐標(biāo)原點(diǎn)SKIPIF1<0為線段SKIPIF1<0的中點(diǎn),SKIPIF1<0,證明：存在定點(diǎn)SKIPIF1<0,使得SKIPIF1<0為定值.6.（2023屆北京市房山區(qū)高三上學(xué)期考試）已知橢圓SKIPIF1<0的長(zhǎng)軸的兩個(gè)端點(diǎn)分別為SKIPIF1<0離心率為SKIPIF1<0．(1)求橢圓C的標(biāo)準(zhǔn)方程；(2)M為橢圓C上除A,B外任意一點(diǎn),直線SKIPIF1<0交直線SKIPIF1<0于點(diǎn)N,點(diǎn)O為坐標(biāo)原點(diǎn),過(guò)點(diǎn)O且與直線SKIPIF1<0垂直的直線記為l,直線SKIPIF1<0交y軸于點(diǎn)P,交直線l于點(diǎn)Q,求證：SKIPIF1<0為定值．7．（2022屆浙江省“數(shù)海漫游”高三上學(xué)期模擬）已知斜率為k的直線l與拋物線y2=4x交于A?B兩點(diǎn),y軸上的點(diǎn)P使得△ABP是等邊三角形.（1）若k>0,證明：點(diǎn)P在y軸正半軸上；（2）當(dāng)SKIPIF1<0取到最大值時(shí),求實(shí)數(shù)k的值.8．（2022屆上海市建平中學(xué)高三上學(xué)期考試）設(shè)實(shí)數(shù)SKIPIF1<0,橢圓D：SKIPIF1<0的右焦點(diǎn)為F,過(guò)F且斜率為k的直線交D于P、Q兩點(diǎn),若線段PQ的中為N,點(diǎn)O是坐標(biāo)原點(diǎn),直線ON交直線SKIPIF1<0于點(diǎn)M．（1）若點(diǎn)P的橫坐標(biāo)為1,求點(diǎn)Q的橫坐標(biāo)；（2）求證：SKIPIF1<0；（3）求SKIPIF1<0的最大值．9．（2022屆江蘇省南京高三上學(xué)期12月聯(lián)考）已知橢圓C：SKIPIF1<00)的離心率為SKIPIF1<0,右頂點(diǎn)為A,過(guò)點(diǎn)B(a,1)的直線l與橢圓C交于不同的兩點(diǎn)M,N,其中點(diǎn)M在第一象限當(dāng)點(diǎn)M,N關(guān)于原點(diǎn)對(duì)稱時(shí),點(diǎn)M的橫坐標(biāo)為SKIPIF1<0．（1）求橢圓C的方程；（2）過(guò)點(diǎn)N作x軸的垂線,與直線AM交于點(diǎn)P,Q為線段NP的中點(diǎn),求直線AQ的斜率,并求線段AQ長(zhǎng)度的最大值．10．（2022屆廣東省華南師范大學(xué)附屬中學(xué)高三上學(xué)期綜合測(cè)試）已知橢圓SKIPIF1<0經(jīng)過(guò)點(diǎn)SKIPIF1<0,離心率為SKIPIF1<0.（1）求橢圓SKIPIF1<0的方程；（2）直線SKIPIF1<0與橢圓SKIPIF1<0相交于SKIPIF1<0兩點(diǎn),求SKIPIF1<0的最大值.11．（2022屆百校聯(lián)盟高三上學(xué)期11月質(zhì)監(jiān)）在平面直角坐標(biāo)系SKIPIF1<0中,動(dòng)點(diǎn)SKIPIF1<0,滿足SKIPIF1<0,記點(diǎn)SKIPIF1<0的軌跡為SKIPIF1<0．（1）請(qǐng)說(shuō)明SKIPIF1<0是什么曲線,并寫(xiě)出它的方程；（2）設(shè)不過(guò)原點(diǎn)SKIPIF1<0且斜率為SKIPIF1<0的直線SKIPIF1<0與SKIPIF1<0交于不同的兩點(diǎn)SKIPIF1<0,SKIPIF1<0,線段SKIPIF1<0的中點(diǎn)為SKIPIF1<0,直線SKIPIF1<0與SKIPIF1<0交于兩點(diǎn)SKIPIF1<0,SKIPIF1<0,請(qǐng)判斷SKIPIF1<0與SKIPIF1<0的關(guān)系,并證明你的結(jié)論．12．（2022屆河南省縣級(jí)示范性高中高三上學(xué)期11月尖子生對(duì)抗賽）已知橢圓SKIPIF1<0：SKIPIF1<0（SKIPIF1<0）與過(guò)原點(diǎn)的直線相交于SKIPIF1<0,SKIPIF1<0兩點(diǎn),上頂點(diǎn)SKIPIF1<0滿足SKIPIF1<0（其中SKIPIF1<0表示直線的概率）．（1）求橢圓SKIPIF1<0的標(biāo)準(zhǔn)方程；（2）若與直線SKIPIF1<0平行且過(guò)橢圓SKIPIF1<0的右焦點(diǎn)SKIPIF1<0的直線SKIPIF1<0