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10 圓錐曲線與向量的交匯（原卷版）

專題10圓錐曲線與向量的交匯一、考情分析平面向量與圓錐曲線的交匯是高考命題的一個(gè)顯著特征,這類試題的常規(guī)形式是用向量形式給出某些條件或結(jié)論,其難點(diǎn)往往不在向量上,對向量部分只需運(yùn)用向量基礎(chǔ)知識(shí)即可實(shí)現(xiàn)相應(yīng)轉(zhuǎn)化.平面向量作為工具可以處理圓錐曲線中的長度、角度、共線、垂直、射影等許多問題,使得這類問題成為高考命題的一個(gè)熱點(diǎn),且時(shí)常出現(xiàn)在解答題中.二、解題秘籍(一)圓錐曲線中常見的向量條件及求解圓錐曲線與向量問題的策略1.設(shè)SKIPIF1<0為直線l的方向向量,若SKIPIF1<0,則l斜率為k；若SKIPIF1<0（m≠0）,則l斜率為SKIPIF1<0；2.A、B、C是平面內(nèi)不重合的三點(diǎn),若有下列條件之一,則A、B、C共線:SKIPIF1<0=1\*GB3①SKIPIF1<0=SKIPIF1<0SKIPIF1<0;=2\*GB3②SKIPIF1<0=SKIPIF1<0SKIPIF1<0+SKIPIF1<0SKIPIF1<0且SKIPIF1<0+SKIPIF1<0=1;=3\*GB3③SKIPIF1<0=(SKIPIF1<0+SKIPIF1<0SKIPIF1<0)/(1+SKIPIF1<0)；=4\*GB3④SKIPIF1<0∥SKIPIF1<0.3.A、B、C是平面內(nèi)不重合的三點(diǎn),若有下列條件之一,則C為線段AB的中點(diǎn):=1\*GB3①SKIPIF1<0=SKIPIF1<0;=2\*GB3②SKIPIF1<0=SKIPIF1<0(SKIPIF1<0+SKIPIF1<0).4.在四邊形ABCD中,若SKIPIF1<0?SKIPIF1<0=0,則ABAC;若∣SKIPIF1<0+SKIPIF1<0∣=∣SKIPIF1<0-SKIPIF1<0∣,則ABAD;若SKIPIF1<0?SKIPIF1<0=SKIPIF1<0?SKIPIF1<0,則ACBD.5.圓錐曲線中涉及向量相等,通常利用橫坐標(biāo)或縱坐標(biāo)相等進(jìn)行轉(zhuǎn)化,涉及向量共線問題,通項(xiàng)利用非零向量SKIPIF1<0共線SKIPIF1<0轉(zhuǎn)化,涉及向量的數(shù)量積,通常利用數(shù)量積的坐標(biāo)運(yùn)算進(jìn)行轉(zhuǎn)化.6.圓錐曲線中兩直線垂直問題,通常轉(zhuǎn)化為兩直線的方向向量的數(shù)量積為零,這樣做可避免討論直線的斜率是否存在.7.圓錐曲線中涉及數(shù)量積問題,通常利用數(shù)量積的坐標(biāo)運(yùn)算把所給條件轉(zhuǎn)化為關(guān)于橫（縱）坐標(biāo)的表達(dá)式.【例1】（2023屆黑龍江省雞西市雞東縣高三上學(xué)期月考）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SKIPIF1<0是否為定值？若是,求出該定值；若不是,請說明理由.【解析】（1）設(shè)SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.因?yàn)镾KIPIF1<0,所以SKIPIF1<0,故SKIPIF1<0的方程為SKIPIF1<0.（2）由題可知直線SKIPIF1<0的斜率一定存在,且不為0,不妨設(shè)直線SKIPIF1<0的方程為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.聯(lián)立方程組SKIPIF1<0,消去SKIPIF1<0整理得SKIPIF1<0,則SKIPIF1<0,整理得SKIPIF1<0.SKIPIF1<0,SKIPIF1<0,則線段SKIPIF1<0的垂直平分線的方程為SKIPIF1<0,令SKIPIF1<0,得SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0.SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0則SKIPIF1<0.故SKIPIF1<0是定值,該定值為SKIPIF1<0.(二)把點(diǎn)共線問題轉(zhuǎn)化為向量共線此類問題通常是把點(diǎn)SKIPIF1<0共線轉(zhuǎn)化為SKIPIF1<0,或點(diǎn)C在直線AB上.【例2】（2022屆新疆昌吉教育體系高三上學(xué)期診斷）已知橢圓SKIPIF1<0的左?右頂點(diǎn)分別為SKIPIF1<0,右焦點(diǎn)為F（1,0）,且橢圓C的離心率為SKIPIF1<0,M,N為橢圓C上任意兩點(diǎn),點(diǎn)P的坐標(biāo)為（4,t）（t≠0）,且滿足SKIPIF1<0.（1）求橢圓C的方程；（2）證明：M,F,N三點(diǎn)共線.【解析】（1）橢圓C的右焦點(diǎn)為SKIPIF1<0,且離心率為SKIPIF1<0,∴a=2,c=1,則b2=a2-c2=3,∴橢圓C的方程為SKIPIF1<0.（2）由（1）知,SKIPIF1<0的坐標(biāo)分別為SKIPIF1<0,設(shè)SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,∵SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0三點(diǎn)共線,SKIPIF1<0三點(diǎn)共線,即SKIPIF1<0,整理得SKIPIF1<0,兩邊平方得SKIPIF1<0,①又M,N在橢圓上,則SKIPIF1<0,代入①并化簡得SKIPIF1<0,又SKIPIF1<0,SKIPIF1<0,∴要證M,F,N三點(diǎn)共線,只需證SKIPIF1<0,即SKIPIF1<0,只需證SKIPIF1<0,整理得SKIPIF1<0,∴M,F,N三點(diǎn)共線.(三)利用向量共線求雙變量的關(guān)系式此類問題一般是給出形如SKIPIF1<0的條件,確定關(guān)于SKIPIF1<0的等式,求解思路是利用兩向量相等橫坐標(biāo)與縱坐標(biāo)分別相等（注意一般情況下橫坐標(biāo)相等與縱坐標(biāo)相等,使用一個(gè)即可,解題時(shí)哪一個(gè)簡單使用哪一個(gè)）,把SKIPIF1<0用其他變量（若點(diǎn)的橫坐標(biāo)或縱坐標(biāo)）表示,再利用題中條件消去其他變量.【例3】（2023屆甘肅省張掖市高三上學(xué)期檢測）橢圓SKIPIF1<0的方程為SKIPIF1<0,過橢圓左焦點(diǎn)SKIPIF1<0且垂直于SKIPIF1<0軸的直線在第二象限與橢圓相交于點(diǎn)SKIPIF1<0,橢圓的右焦點(diǎn)為SKIPIF1<0,已知SKIPIF1<0,橢圓過點(diǎn)SKIPIF1<0．(1)求橢圓SKIPIF1<0的標(biāo)準(zhǔn)方程；(2)過橢圓SKIPIF1<0的右焦點(diǎn)SKIPIF1<0作直線SKIPIF1<0交橢圓SKIPIF1<0于SKIPIF1<0兩點(diǎn),交SKIPIF1<0軸于SKIPIF1<0點(diǎn),若SKIPIF1<0,SKIPIF1<0,求證：SKIPIF1<0為定值．【解析】（1）依題可知：SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0又∵橢圓SKIPIF1<0過點(diǎn)SKIPIF1<0,則SKIPIF1<0聯(lián)立SKIPIF1<0可得SKIPIF1<0,橢圓SKIPIF1<0的標(biāo)準(zhǔn)方程為SKIPIF1<0．（2）設(shè)點(diǎn)SKIPIF1<0、SKIPIF1<0,SKIPIF1<0,由題意可知,直線SKIPIF1<0的斜率存在,可設(shè)直線SKIPIF1<0的方程為SKIPIF1<0,聯(lián)立SKIPIF1<0,可得SKIPIF1<0,由于點(diǎn)SKIPIF1<0在橢圓SKIPIF1<0的內(nèi)部,直線SKIPIF1<0與橢圓SKIPIF1<0必有兩個(gè)交點(diǎn),由韋達(dá)定理可得SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0．(四)利用向量加法的幾何意義構(gòu)造平行四邊形若點(diǎn)SKIPIF1<0滿足SKIPIF1<0,則四邊形ABCD是平行四邊形,涉及圓錐曲線中的平行四邊形要注意對邊長度相等、斜率相等,兩對角線中點(diǎn)為同一個(gè)點(diǎn)等條件的應(yīng)用.【例4】（2023屆四川省廣安市岳池縣高三上學(xué)期10月月考）已知橢圓SKIPIF1<0經(jīng)過點(diǎn)SKIPIF1<0,左焦點(diǎn)SKIPIF1<0.(1)求橢圓SKIPIF1<0的方程；(2)過點(diǎn)SKIPIF1<0作直線SKIPIF1<0與橢圓SKIPIF1<0交于SKIPIF1<0兩點(diǎn),點(diǎn)SKIPIF1<0滿足SKIPIF1<0（SKIPIF1<0為原點(diǎn)）,求四邊形SKIPIF1<0面積的最大值.【解析】（1）設(shè)橢圓的焦距為SKIPIF1<0,則SKIPIF1<0,又因?yàn)闄E圓經(jīng)過點(diǎn)SKIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以橢圓SKIPIF1<0的方程為SKIPIF1<0.（2）因?yàn)镾KIPIF1<0,所以四邊形SKIPIF1<0為平行四邊形,當(dāng)直線SKIPIF1<0的斜率不存在時(shí),顯然不符合題意；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,SKIPIF1<0與橢圓交于SKIPIF1<0,SKIPIF1<0兩點(diǎn),由SKIPIF1<0.由SKIPIF1<0SKIPIF1<0,SKIPIF1<0,SKIPIF1<0SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0（由上式知SKIPIF1<0）,SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí)取等號(hào).∴當(dāng)SKIPIF1<0時(shí),平行四邊形SKIPIF1<0的面積最大值為2.(五)把向量的數(shù)量積轉(zhuǎn)化為代數(shù)式若圓錐曲線問題有用向量數(shù)量積給出的條件,通常是利用向量數(shù)量積的坐標(biāo)運(yùn)算進(jìn)行轉(zhuǎn)化.【例5】（2023屆廣東省荔灣區(qū)高三上學(xué)期10月調(diào)研）已知雙曲線SKIPIF1<0的右焦點(diǎn)為SKIPIF1<0為坐標(biāo)原點(diǎn),雙曲線SKIPIF1<0的兩條漸近線的夾角為SKIPIF1<0．(1)求雙曲線SKIPIF1<0的方程；(2)過點(diǎn)SKIPIF1<0作直線SKIPIF1<0交SKIPIF1<0于SKIPIF1<0兩點(diǎn),在SKIPIF1<0軸上是否存在定點(diǎn)SKIPIF1<0,使SKIPIF1<0為定值？若存在,求出定點(diǎn)SKIPIF1<0的坐標(biāo)及這個(gè)定值；若不存在,說明理由．【解析】（1）雙曲線SKIPIF1<0的漸近線為SKIPIF1<0,又SKIPIF1<0,SKIPIF1<0,故其漸近線SKIPIF1<0的傾斜角小于SKIPIF1<0,而雙曲線SKIPIF1<0的兩條漸近線的夾角為SKIPIF1<0,則漸近線的SKIPIF1<0的傾斜角為SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0．又SKIPIF1<0,則SKIPIF1<0．所以雙曲線SKIPIF1<0的方程是SKIPIF1<0．（2）當(dāng)直線SKIPIF1<0不與SKIPIF1<0軸重合時(shí),設(shè)直線SKIPIF1<0的方程為SKIPIF1<0,代入SKIPIF1<0,得SKIPIF1<0,即SKIPIF1<0．設(shè)點(diǎn)SKIPIF1<0,則SKIPIF1<0．設(shè)點(diǎn)SKIPIF1<0,則SKIPIF1<0SKIPIF1<0SKIPIF1<0令SKIPIF1<0,得SKIPIF1<0,此時(shí)SKIPIF1<0．當(dāng)直線SKIPIF1<0與SKIPIF1<0軸重合時(shí),則點(diǎn)SKIPIF1<0為雙曲線的兩頂點(diǎn),不妨設(shè)點(diǎn)SKIPIF1<0．對于點(diǎn)SKIPIF1<0．所以存在定點(diǎn)SKIPIF1<0,使SKIPIF1<0為定值．(六)把垂直問題轉(zhuǎn)化為向量的數(shù)量積為零求解圓錐曲線中的垂直問題,通?？赊D(zhuǎn)化為向量的數(shù)量積為零,然后利用向量數(shù)量積的坐標(biāo)運(yùn)算進(jìn)行轉(zhuǎn)化,這種轉(zhuǎn)化可避免討論直線的斜率是否存在.【例6】已知橢圓SKIPIF1<0的右焦點(diǎn)為SKIPIF1<0,橢圓SKIPIF1<0上的點(diǎn)到SKIPIF1<0的距離的最大值和最小值分別為SKIPIF1<0和SKIPIF1<0．（1）求橢圓SKIPIF1<0的標(biāo)準(zhǔn)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),是否存在正數(shù)SKIPIF1<0,使得SKIPIF1<0？若存在,求出SKIPIF1<0的值；若不存在,請說明理由．【解析】（1）由題意可得,SKIPIF1<0,解得SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,所以橢圓方程為SKIPIF1<0；（2）假設(shè)存在正數(shù)SKIPIF1<0,使得SKIPIF1<0,即使得SKIPIF1<0,當(dāng)直線SKIPIF1<0的斜率不存在時(shí),設(shè)直線SKIPIF1<0的方程為SKIPIF1<0,可得SKIPIF1<0,SKIPIF1<0,因?yàn)镾KIPIF1<0,則有SKIPIF1<0,解得SKIPIF1<0,又直線SKIPIF1<0為圓SKIPIF1<0的切線,所以SKIPIF1<0；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,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,聯(lián)立SKIPIF1<0,可得SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,且SKIPIF1<0,所以SKIPIF1<0,因?yàn)镾KIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,整理可得SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,因?yàn)橹本€SKIPIF1<0為圓SKIPIF1<0的切線,故原點(diǎn)SKIPIF1<0到SKIPIF1<0的距離為SKIPIF1<0,所以存在正數(shù)SKIPIF1<0,使得SKIPIF1<0．三、跟蹤檢測1.（2023屆重慶市第八中學(xué)校高三上學(xué)期月考）已知雙曲線E：SKIPIF1<0（SKIPIF1<0,SKIPIF1<0）一個(gè)頂點(diǎn)為SKIPIF1<0,直線l過點(diǎn)SKIPIF1<0交雙曲線右支于M,N兩點(diǎn),記SKIPIF1<0,SKIPIF1<0,SKIPIF1<0的面積分別為S,SKIPIF1<0,SKIPIF1<0.當(dāng)l與x軸垂直時(shí),SKIPIF1<0的值為SKIPIF1<0.(1)求雙曲線E的標(biāo)準(zhǔn)方程；(2)若l交y軸于點(diǎn)P,SKIPIF1<0,SKIPIF1<0,求證：SKIPIF1<0為定值；(3)在（2）的條件下,若SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),求實(shí)數(shù)m的取值范圍.2.（2023屆江蘇省連云港市高三上學(xué)期10月聯(lián)考）已知橢圓中有兩頂點(diǎn)為SKIPIF1<0,SKIPIF1<0,一個(gè)焦點(diǎn)為SKIPIF1<0.(1)若直線SKIPIF1<0過點(diǎn)SKIPIF1<0且與橢圓交于SKIPIF1<0,SKIPIF1<0兩點(diǎn),當(dāng)SKIPIF1<0時(shí),求直線SKIPIF1<0的方程；(2)若直線SKIPIF1<0過點(diǎn)SKIPIF1<0且與橢圓交于SKIPIF1<0,SKIPIF1<0兩點(diǎn),并與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0,直線SKIPIF1<0與直線SKIPIF1<0交于點(diǎn)SKIPIF1<0,當(dāng)點(diǎn)SKIPIF1<0異SKIPIF1<0,SKIPIF1<0兩點(diǎn)時(shí),試問SKIPIF1<0是否是定值？若是,請求出此定值,若不是,請說明理由.3.（2023屆四川省成都市郫都區(qū)高三上學(xué)期檢測）已知橢圓SKIPIF1<0的離心率為SKIPIF1<0,短軸長為4．(1)求橢圓C的方程；(2)若過點(diǎn)SKIPIF1<0的直線交橢圓C于A,B兩點(diǎn),求SKIPIF1<0的取值范圍．4.（2023屆江蘇省南通市如皋市高三上學(xué)期9月診斷測試）已知點(diǎn)SKIPIF1<0分別是橢圓SKIPIF1<0的左、右頂點(diǎn),過SKIPIF1<0的右焦點(diǎn)SKIPIF1<0作直線SKIPIF1<0交SKIPIF1<0于SKIPIF1<0兩點(diǎn),(1)設(shè)直線SKIPIF1<0的斜率分別為SKIPIF1<0,求SKIPIF1<0和SKIPIF1<0的值；(2)若直線SKIPIF1<0分別交橢圓SKIPIF1<0的右準(zhǔn)線于SKIPIF1<0兩點(diǎn),證明：以SKIPIF1<0為直徑的圓經(jīng)過定點(diǎn).5.（2023屆湖南省部分校高三上學(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SKIPIF1<0上.(1)求雙曲線SKIPIF1<0的方程.(2)設(shè)過點(diǎn)SKIPIF1<0的直線SKIPIF1<0與雙曲線SKIPIF1<0交于SKIPIF1<0兩點(diǎn),問在SKIPIF1<0軸上是否存在定點(diǎn)SKIPIF1<0,使得SKIPIF1<0為常數(shù)？若存在,求出點(diǎn)SKIPIF1<0的坐標(biāo)以及該常數(shù)的值；若不存在,請說明理由.6.（2023屆廣東省茂名市高三上學(xué)期9月聯(lián)考）如圖,平面直角坐標(biāo)系SKIPIF1<0中,點(diǎn)SKIPIF1<0為SKIPIF1<0軸上的一個(gè)動(dòng)點(diǎn),動(dòng)點(diǎn)SKIPIF1<0滿足SKIPIF1<0,又點(diǎn)SKIPIF1<0滿足SKIPIF1<0.(1)求動(dòng)點(diǎn)SKIPIF1<0的軌跡SKIPIF1<0的方程；(2)過曲線SKIPIF1<0上的點(diǎn)SKIPIF1<0（SKIPIF1<0）的直線SKIPIF1<0與SKIPIF1<0,SKIPIF1<0軸的交點(diǎn)分別為SKIPIF1<0和SKIPIF1<0,且SKIPIF1<0,過原點(diǎn)SKIPIF1<0的直線與SKIPIF1<0平行,且與曲線SKIPIF1<0交于SKIPIF1<0、SKIPIF1<0兩點(diǎn),求SKIPIF1<0面積的最大值.7.（2023屆福建師范大學(xué)附屬中學(xué)高三上學(xué)期月考）在平面直角坐標(biāo)系SKIPIF1<0中,設(shè)點(diǎn)SKIPIF1<0,點(diǎn)SKIPIF1<0與SKIPIF1<0兩點(diǎn)的距離之和為SKIPIF1<0為一動(dòng)點(diǎn),點(diǎn)SKIPIF1<0滿足向量關(guān)系式：SKIPIF1<0．(1)求點(diǎn)SKIPIF1<0的軌跡方程SKIPIF1<0；(2)設(shè)SKIPIF1<0與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0(SKIPIF1<0在SKIPIF1<0的左側(cè)),點(diǎn)SKIPIF1<0為SKIPIF1<0上一動(dòng)點(diǎn)(且不與SKIPIF1<0重合)．設(shè)直線SKIPIF1<0軸與直線SKIPIF1<0分別交于點(diǎn)SKIPIF1<0,取SKIPIF1<0,連接SKIPIF1<0,證明：SKIPIF1<0為SKIPIF1<0的角平分線．8.（2023屆山西省山西大學(xué)附屬中學(xué)校高三上學(xué)期9月診斷）如圖,橢圓SKIPIF1<0：SKIPIF1<0（SKIPIF1<0,SKIPIF1<0,SKIPIF1<0是橢圓SKIPIF1<0的左焦點(diǎn),SKIPIF1<0是橢圓SKIPIF1<0的左頂點(diǎn),SKIPIF1<0是橢圓SKIPIF1<0的上頂點(diǎn),且SKIPIF1<0,點(diǎn)SKIPIF1<0SKIPIF1<0是長軸上的任一定點(diǎn),過SKIPIF1<0點(diǎn)的任一直線SKIPIF1<0交橢圓SKIPIF1<0于SKIPIF1<0兩點(diǎn).(1)求橢圓SKIPIF1<0的方程；(2)是否存在定點(diǎn)SKIPIF1<0,使得SKIPIF1<0為定值,若存在,試求出定點(diǎn)SKIPIF1<0的坐標(biāo),并求出此定值；若不存在,請說明理由.9．（2023屆北京市第四中學(xué)高三上學(xué)期開學(xué)測試）已知中心在原點(diǎn),焦點(diǎn)在SKIPIF1<0軸上的橢圓SKIPIF1<0過點(diǎn)SKIPIF1<0,離心率為SKIPIF1<0,點(diǎn)SKIPIF1<0為其右頂點(diǎn).過點(diǎn)SKIPIF1<0作直線SKIPIF1<0與橢圓SKIPIF1<0相交于SKIPIF1<0、SKIPIF1<0兩點(diǎn),直線SKIPIF1<0、SKIPIF1<0與直線SKIPIF1<0分別交于點(diǎn)SKIPIF1<0、SKIPIF1<0.(1)求橢圓SKIPIF1<0的方程；(2)求SKIPIF1<0的取值范圍.10．（2023屆湖北省“宜荊荊恩”高三上學(xué)期考試）已知雙曲線SKIPIF1<0與雙曲線SKIPIF1<0有相同的漸近線,且過點(diǎn)SKIPIF1<0.(1)求雙曲線SKIPIF1<0的標(biāo)準(zhǔn)方程；(2)已知SKIPIF1<0是雙曲線SKIPIF1<0上不同于SKIPIF1<0的兩點(diǎn),且SKIPIF1<0于SKIPIF1<0,證明：存在定點(diǎn)SKIPIF1<0,使SKIPIF1<0為定值.11．（2023屆四川省達(dá)州市開江縣高三上學(xué)期考試）已知橢圓SKIPIF1<0?為橢圓SKIPIF1<0?的左、右焦點(diǎn),過點(diǎn)SKIPIF1<0?的任意直線SKIPIF1<0?交橢圓SKIPIF1<0?于SKIPIF1<0、SKIPIF1<0?兩點(diǎn),且SKIPIF1<0的周長為8,橢圓SKIPIF1<0?的離心率為SKIPIF1<0?.(1)橢圓SKIPIF1<0?的方程；(2)若SKIPIF1<0?為橢圓SKIPIF1<0?上的任一點(diǎn),SKIPIF1<0?為過焦點(diǎn)SKIPIF1<0?的弦,且SKIPIF1<0?,求SKIPIF1<0?的值.12．（2022屆上海市普陀區(qū)高三一模）已知點(diǎn)SKIPIF1<0與定點(diǎn)SKIPIF1<0的距離是點(diǎn)SKIPIF1<0到直線SKIPIF1<0距離的SKIPIF1<0倍,設(shè)點(diǎn)SKIPIF1<0的軌跡為曲線SKIPIF1<0,直線SKIPIF1<0與SKIPIF1<0交于SKIPIF1<0、SKIPIF1<0兩點(diǎn),點(diǎn)SKIPIF1<0是線段SKIPIF1<0的中點(diǎn),SKIPIF1<0、SKIPIF1<0是SKIPIF1<0上關(guān)于原點(diǎn)SKIPIF1<0對稱的兩點(diǎn),且SKIPIF1<0.（1）求曲線SKIPIF1<0的方程；（2）當(dāng)SKIPIF1<0時(shí),求直線SKIPIF1<0的方程；（3）當(dāng)四邊形SKIPIF1<0的面積SKIPIF1<0時(shí),求SKIPIF1<0的值.13．(2022屆內(nèi)蒙古赤峰市高三上學(xué)期11月聯(lián)考)已知橢圓SKIPIF1<0的焦點(diǎn)恰為橢圓SKIPIF1<0長軸的端點(diǎn),且SKIPIF1<0的短軸長為2（1）求橢圓SKIPIF1<0的方程．（2）若直線SKIPIF1<0與直線SKIPIF1<0平行,且SKIPIF1<0與SKIPIF1<0交于SKIPIF1<0,SKIPIF1<0兩點(diǎn),SKIPIF1<0,求SKIPIF1<0的最小值．14．（2022屆遼寧省大連市高三上學(xué)期期中）在平面直角坐標(biāo)系SKIPIF1<0中,點(diǎn)SKIPIF1<0,SKIPIF1<0的坐標(biāo)分別為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0是動(dòng)點(diǎn),且直線SKIPIF1<0與SKIPIF1<0的斜率之積等于SKIPIF1<0.（1）求動(dòng)點(diǎn)SKIPIF1<0的軌跡SKIPIF1<0的方程；（2）已知直線SKIPIF1<0與橢圓：SKIPIF1<0相交于SKIPIF1<0,SKIPIF1<0兩點(diǎn),與SKIPIF1<0軸交于點(diǎn)SK