新高考數(shù)學(xué)一輪復(fù)習(xí)章節(jié)專題模擬卷第八章 解析幾何（直線與圓、圓錐曲線）（解析卷）

解析幾何本試卷22小題，滿分150分?？荚囉脮r(shí)120分鐘一、單項(xiàng)選擇題：本題共8小題，每小題5分，共40分。在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的。1．（2023·廣東·高三統(tǒng)考模擬預(yù)測(cè)）設(shè)SKIPIF1<0，則“SKIPIF1<0”是“直線SKIPIF1<0與直線SKIPIF1<0平行”的（

）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件【答案】A【解析】若直線SKIPIF1<0與直線SKIPIF1<0平行，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，經(jīng)檢驗(yàn)SKIPIF1<0或SKIPIF1<0時(shí)兩直線平行．故“SKIPIF1<0”能得到“直線SKIPIF1<0與直線SKIPIF1<0平行”，但是“直線SKIPIF1<0與直線SKIPIF1<0平行”不能得到“SKIPIF1<0”故選：A2.(2023·南京模擬)已知橢圓的兩個(gè)焦點(diǎn)分別為F1(0,2),F2(0，－2)，P為橢圓上任意一點(diǎn)，若|F1F2|是|PF1|，|PF2|的等差中項(xiàng)，則此橢圓的標(biāo)準(zhǔn)方程為()A.eq\f(x2,64)＋eq\f(y2,60)＝1 B.eq\f(y2,64)＋eq\f(x2,60)＝1C.eq\f(x2,16)＋eq\f(y2,12)＝1 D.eq\f(y2,16)＋eq\f(x2,12)＝1【答案】D【解析】由題意|PF1|＋|PF2|＝2|F1F2|＝8＝2a，故a＝4，又c＝2，則b＝2eq\r(3)，焦點(diǎn)在y軸上，故橢圓的標(biāo)準(zhǔn)方程為eq\f(y2,16)＋eq\f(x2,12)＝1.3．（2023·廣東江門·統(tǒng)考模擬預(yù)測(cè)）若直線SKIPIF1<0與圓SKIPIF1<0相交于P，Q兩點(diǎn)，且SKIPIF1<0（其中O為坐標(biāo)原點(diǎn)），則b的值為（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】SKIPIF1<0，圓的半徑為1，SKIPIF1<0，圓心SKIPIF1<0到直線SKIPIF1<0的距離SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0.故選：C4．(2023·昆明模擬)已知橢圓eq\f(x2,4)＋eq\f(y2,3)＝1的兩個(gè)焦點(diǎn)為F1，F(xiàn)2，過(guò)F2的直線交橢圓于M，N兩點(diǎn)，則△F1MN的周長(zhǎng)為()A．2B．4C．6D．8【答案】D【解析】由eq\f(x2,4)＋eq\f(y2,3)＝1得a＝2.因?yàn)镸，N是橢圓上的點(diǎn)，F(xiàn)1，F(xiàn)2是橢圓的焦點(diǎn)，所以|MF1|＋|MF2|＝2a，|NF1|＋|NF2|＝2a，因此△F1MN的周長(zhǎng)為|MF1|＋|MN|＋|NF1|＝|MF1|＋|MF2|＋|NF2|＋|NF1|＝2a＋2a＝4a＝8.5．（2023·湖南長(zhǎng)沙·長(zhǎng)沙市明德中學(xué)?？既＃┮阎獟佄锞€SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，準(zhǔn)線為SKIPIF1<0，SKIPIF1<0為SKIPIF1<0上一點(diǎn)，SKIPIF1<0，垂足為SKIPIF1<0，SKIPIF1<0與SKIPIF1<0軸交點(diǎn)為SKIPIF1<0，若SKIPIF1<0，且SKIPIF1<0的面積為SKIPIF1<0，則SKIPIF1<0的方程為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】由拋物線定義知SKIPIF1<0，所以SKIPIF1<0為等邊三角形，SKIPIF1<0為SKIPIF1<0的中點(diǎn)，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0的面積SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0的方程為SKIPIF1<0.故選：A.6．（2023·江蘇·統(tǒng)考三模）已知F為橢圓C：SKIPIF1<0的右焦點(diǎn)，P為C上一點(diǎn)，Q為圓M：SKIPIF1<0上一點(diǎn)，則PQ＋PF的最大值為（

）A．3 B．6C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】圓M：SKIPIF1<0的圓心為SKIPIF1<0，設(shè)橢圓的左焦點(diǎn)為SKIPIF1<0，如下圖，由橢圓的定義知，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0三點(diǎn)在一條直線上時(shí)取等，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.故選：D．7．（2023·浙江·統(tǒng)考二模）已知SKIPIF1<0是圓SKIPIF1<0上一點(diǎn)，SKIPIF1<0是圓SKIPIF1<0的直徑，弦SKIPIF1<0的中點(diǎn)為SKIPIF1<0.若點(diǎn)SKIPIF1<0在第一象限，直線SKIPIF1<0、SKIPIF1<0的斜率之和為0，則直線SKIPIF1<0的斜率是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由題可得圓的方程，設(shè)直線SKIPIF1<0的斜率為SKIPIF1<0，則直線SKIPIF1<0的方程為SKIPIF1<0，代入圓的方程可得SKIPIF1<0的坐標(biāo)，從而可得SKIPIF1<0的坐標(biāo)，于是根據(jù)斜率關(guān)系可解得SKIPIF1<0的值，由于點(diǎn)SKIPIF1<0在第一象限，對(duì)SKIPIF1<0的值進(jìn)行取舍，即可得所求.【詳解】已知SKIPIF1<0是圓SKIPIF1<0上一點(diǎn)，所以SKIPIF1<0設(shè)直線SKIPIF1<0的斜率為SKIPIF1<0，則直線SKIPIF1<0的方程為SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0恒成立，所以SKIPIF1<0由于SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，由于SKIPIF1<0是圓SKIPIF1<0的直徑，所以SKIPIF1<0，則弦SKIPIF1<0的中點(diǎn)為SKIPIF1<0坐標(biāo)為SKIPIF1<0因?yàn)橹本€SKIPIF1<0、SKIPIF1<0的斜率之和為0，所以SKIPIF1<0，整理得SKIPIF1<0解得SKIPIF1<0或SKIPIF1<0，又點(diǎn)SKIPIF1<0在第一象限，所以SKIPIF1<0，故SKIPIF1<0，即直線SKIPIF1<0的斜率是SKIPIF1<0.故選：C.8.(2022·濟(jì)南模擬)已知拋物線C：y2＝4x，圓F：(x－1)2＋y2＝1，直線l：y＝k(x－1)(k≠0)自上而下順次與上述兩曲線交于M1，M2，M3，M4四點(diǎn)，則下列各式結(jié)果為定值的是()A．|M1M2|·|M3M4| B．|FM1|·|FM4|C．|M1M3|·|M2M4| D．|FM1|·|M1M2|【答案】A【解析】如圖，分別設(shè)M1，M2，M3，M4四點(diǎn)的橫坐標(biāo)為x1，x2，x3，x4，由y2＝4x得焦點(diǎn)F(1,0)，準(zhǔn)線l0：x＝－1，由定義得，|M1F|＝x1＋1，又|M1F|＝|M1M2|＋1，所以|M1M2|＝x1，同理|M3M4|＝x4，由eq\b\lc\{\rc\(\a\vs4\al\co1(y＝kx－1，,y2＝4x，))消去y，整理得k2x2－(2k2＋4)x＋k2＝0(k≠0)，設(shè)M1(x1，y1)，M4(x4，y4)，則x1x4＝1，即|M1M2|·|M3M4|＝1.二、多項(xiàng)選擇題：本題共4小題，每小題5分，共20分。在每小題給出的選項(xiàng)中，有多項(xiàng)符合題目要求。全部選對(duì)的得5分，部分選對(duì)的得2分，有選錯(cuò)的得0分。9．（2023·廣東肇慶·統(tǒng)考一模）已知圓SKIPIF1<0，直線SKIPIF1<0，則（

）A．直線SKIPIF1<0過(guò)定點(diǎn)SKIPIF1<0B．直線SKIPIF1<0與圓SKIPIF1<0可能相離C．圓SKIPIF1<0被SKIPIF1<0軸截得的弦長(zhǎng)為SKIPIF1<0D．圓SKIPIF1<0被直線SKIPIF1<0截得的弦長(zhǎng)最短時(shí)，直線SKIPIF1<0的方程為SKIPIF1<0【答案】AC【解析】直線SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，即l恒過(guò)定點(diǎn)SKIPIF1<0，故A正確；點(diǎn)SKIPIF1<0與圓心SKIPIF1<0的距離SKIPIF1<0，故直線l與圓C恒相交，故B錯(cuò)誤；令SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，故圓C被y軸截得的弦長(zhǎng)為SKIPIF1<0，故C正確；要使直線l被圓C截得弦長(zhǎng)最短，只需SKIPIF1<0與圓心SKIPIF1<0連線垂直于直線SKIPIF1<0，所以直線l的斜率SKIPIF1<0，可得SKIPIF1<0，故直線l為SKIPIF1<0，故D錯(cuò)誤．故選：AC．10．（2023·安徽馬鞍山·統(tǒng)考三模）已知拋物線SKIPIF1<0：SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，點(diǎn)SKIPIF1<0為坐標(biāo)原點(diǎn)，點(diǎn)SKIPIF1<0在拋物線上，直線SKIPIF1<0與拋物線SKIPIF1<0交于點(diǎn)SKIPIF1<0，則（

）A．SKIPIF1<0的準(zhǔn)線方程為SKIPIF1<0 B．SKIPIF1<0C．直線SKIPIF1<0的斜率為SKIPIF1<0 D．SKIPIF1<0【答案】CD【詳解】由題意可知，SKIPIF1<0，得SKIPIF1<0，則拋物線方程為SKIPIF1<0，所以拋物線的準(zhǔn)線方程為SKIPIF1<0，故A錯(cuò)誤；拋物線的焦點(diǎn)SKIPIF1<0，SKIPIF1<0，則直線SKIPIF1<0的方程為SKIPIF1<0，與拋物線方程聯(lián)立SKIPIF1<0，得SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故B錯(cuò)誤，C正確；SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故D正確.故選：CD11.．(2023·湖北四地聯(lián)考)已知橢圓C：eq\f(x2,a2)＋eq\f(y2,b2)＝1(a>b>0)的左、右焦點(diǎn)分別為F1，F(xiàn)2，長(zhǎng)軸長(zhǎng)為4，點(diǎn)P(eq\r(2)，1)在橢圓C外，點(diǎn)Q在橢圓C上，則()A．橢圓C的離心率的取值范圍是SKIPIF1<0B．當(dāng)橢圓C的離心率為eq\f(\r(3),2)時(shí)，|QF1|的取值范圍是[2－eq\r(3)，2＋eq\r(3)]C．存在點(diǎn)Q使得eq\o(QF1,\s\up6(→))·eq\o(QF2,\s\up6(→))＝0D.eq\f(1,|QF1|)＋eq\f(1,|QF2|)的最小值為1【答案】BCD【解析】由題意得a＝2，又點(diǎn)P(eq\r(2)，1)在橢圓C外，則eq\f(2,4)＋eq\f(1,b2)>1，解得b<eq\r(2)，所以橢圓C的離心率e＝eq\f(c,a)＝eq\f(\r(4－b2),2)>eq\f(\r(2),2)，即橢圓C的離心率的取值范圍是SKIPIF1<0，故A不正確；當(dāng)e＝eq\f(\r(3),2)時(shí)，c＝eq\r(3)，b＝eq\r(a2－c2)＝1，所以|QF1|的取值范圍是[a－c，a＋c]，即[2－eq\r(3)，2＋eq\r(3)]，故B正確；設(shè)橢圓的上頂點(diǎn)為A(0，b)，F(xiàn)1(－c,0)，F(xiàn)2(c,0)，由于eq\o(AF1,\s\up6(→))·eq\o(AF2,\s\up6(→))＝b2－c2＝2b2－a2<0，所以存在點(diǎn)Q使得eq\o(QF1,\s\up6(→))·eq\o(QF2,\s\up6(→))＝0，故C正確；(|QF1|＋|QF2|)SKIPIF1<0＝2＋eq\f(|QF2|,|QF1|)＋eq\f(|QF1|,|QF2|)≥2＋2＝4，當(dāng)且僅當(dāng)|QF1|＝|QF2|＝2時(shí)，等號(hào)成立，又|QF1|＋|QF2|＝4，所以eq\f(1,|QF1|)＋eq\f(1,|QF2|)≥1，故D正確．12．（2023·湖南邵陽(yáng)·統(tǒng)考三模）已知雙曲線CSKIPIF1<0的左?右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，雙曲線具有如下光學(xué)性質(zhì)：從右焦點(diǎn)SKIPIF1<0發(fā)出的光線m交雙曲線右支于點(diǎn)P，經(jīng)雙曲線反射后，反射光線n的反向延長(zhǎng)線過(guò)左焦點(diǎn)SKIPIF1<0，如圖所示.若雙曲線C的一條漸近線的方程為SKIPIF1<0，則下列結(jié)論正確的有（

）A．雙曲線C的方程為SKIPIF1<0B．若SKIPIF1<0，則SKIPIF1<0C．若射線n所在直線的斜率為k，則SKIPIF1<0D．當(dāng)n過(guò)點(diǎn)M（8，5）時(shí)，光由SKIPIF1<0所經(jīng)過(guò)的路程為10【答案】AC【詳解】對(duì)于A，由題意可知，SKIPIF1<0因?yàn)殡p曲線C的一條漸近線的方程為SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以雙曲線的方程為SKIPIF1<0故A正確；對(duì)于B，由SKIPIF1<0SKIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，由勾股定理及雙曲線的定義知，SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，故B錯(cuò)誤；對(duì)于C，由題意可知，雙曲線的漸近線方程為SKIPIF1<0，由雙曲線的性質(zhì)可得射線SKIPIF1<0所在直線的斜率范圍為SKIPIF1<0，故C正確；對(duì)于D，由題意可知，SKIPIF1<0，當(dāng)SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0時(shí)，由雙曲線定義可得光由SKIPIF1<0所經(jīng)過(guò)的路程為SKIPIF1<0，故D錯(cuò)誤.故選：AC.三、填空題：本大題共4小題，每小題5分，共20分。13．（2023·浙江臺(tái)州·統(tǒng)考二模）已知橢圓SKIPIF1<0經(jīng)過(guò)點(diǎn)SKIPIF1<0和SKIPIF1<0，則橢圓SKIPIF1<0的離心率為_(kāi)__________.【答案】SKIPIF1<0/0.5【分析】通過(guò)已知兩個(gè)點(diǎn)求出橢圓方程即可得到離心率.【詳解】將兩個(gè)點(diǎn)代入橢圓方程得：SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0.故答案為：SKIPIF1<014．（2023·浙江·統(tǒng)考二模）已知圓SKIPIF1<0，若SKIPIF1<0被兩坐標(biāo)軸截得的弦長(zhǎng)相等，則SKIPIF1<0__________.【答案】SKIPIF1<0/SKIPIF1<0【分析】SKIPIF1<0被兩坐標(biāo)軸截得的弦長(zhǎng)相等，則圓心SKIPIF1<0到兩坐標(biāo)軸的距離相等，即圓心的橫縱坐標(biāo)的絕對(duì)值相等可得答案.【詳解】圓的弦長(zhǎng)為SKIPIF1<0（SKIPIF1<0為圓的半徑，SKIPIF1<0為圓心到弦的距離），若SKIPIF1<0被兩坐標(biāo)軸截得的弦長(zhǎng)相等，則圓心SKIPIF1<0到兩坐標(biāo)軸的距離相等，即圓心的橫縱坐標(biāo)的絕對(duì)值相等，即SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<0.15.(2023·長(zhǎng)沙模擬)已知拋物線C：y2＝16x，傾斜角為eq\f(π,6)的直線l過(guò)焦點(diǎn)F交拋物線于A，B兩點(diǎn)，O為坐標(biāo)原點(diǎn)，則△ABO的面積為_(kāi)_______．【答案】64【解析】方法一(常規(guī)解法)依題意，拋物線C：y2＝16x的焦點(diǎn)為F(4,0)，直線l的方程為x＝eq\r(3)y＋4.由eq\b\lc\{\rc\(\a\vs4\al\co1(x＝\r(3)y＋4，,y2＝16x，))消去x，整理得y2－16eq\r(3)y－64＝0.設(shè)A(x1，y1)，B(x2，y2)，則y1＋y2＝16eq\r(3)，y1y2＝－64.S△OAB＝eq\f(1,2)|y1－y2|·|OF|＝2eq\r(y1＋y22－4y1y2)＝2eq\r(16\r(3)2－4×－64)＝64.方法二(活用結(jié)論)依題意，拋物線y2＝16x，p＝8.又l的傾斜角α＝eq\f(π,6).所以S△OAB＝eq\f(p2,2sinα)＝eq\f(82,2sin

\f(π,6))＝64.16．（2023·遼寧大連·統(tǒng)考三模）已知SKIPIF1<0為坐標(biāo)原點(diǎn)，SKIPIF1<0是雙曲線SKIPIF1<0的左?右焦點(diǎn)，雙曲線SKIPIF1<0上一點(diǎn)SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0，則雙曲線SKIPIF1<0的漸近線方程為_(kāi)_________.點(diǎn)A是雙曲線SKIPIF1<0上一定點(diǎn)，過(guò)點(diǎn)SKIPIF1<0的動(dòng)直線SKIPIF1<0與雙曲線SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，SKIPIF1<0為定值SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)實(shí)數(shù)SKIPIF1<0的值為_(kāi)_________.【答案】SKIPIF1<0SKIPIF1<0【詳解】（1）根據(jù)SKIPIF1<0，取SKIPIF1<0的中點(diǎn)SKIPIF1<0，易知SKIPIF1<0，可知SKIPIF1<0，SKIPIF1<0，即△SKIPIF1<0為直角三角形.設(shè)SKIPIF1<0，依題意有SKIPIF1<0，解得SKIPIF1<0，根據(jù)勾股定理得SKIPIF1<0，解得SKIPIF1<0，故雙曲線為等軸雙曲線，漸近線為SKIPIF1<0.（2）當(dāng)SKIPIF1<0時(shí)，雙曲線SKIPIF1<0，設(shè)直線SKIPIF1<0，聯(lián)立方程組SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，因?yàn)镾KIPIF1<0為定值SKIPIF1<0，所以法一：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0法二：SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<0，SKIPIF1<0四、解答題：本大題共6小題，共70分，請(qǐng)?jiān)诖痤}卡指定區(qū)域內(nèi)作答，解答時(shí)應(yīng)寫出文字說(shuō)明、證明過(guò)程或演算步驟。17.(2023·衡水模擬)已知橢圓C：eq\f(x2,a2)＋eq\f(y2,b2)＝1(a>b>0)的左、右焦點(diǎn)分別為F1，F(xiàn)2，離心率為eq\f(\r(2),2)，短軸頂點(diǎn)分別為M，N，四邊形MF1NF2的面積為32.(1)求橢圓C的標(biāo)準(zhǔn)方程；(2)直線l交橢圓C于A，B兩點(diǎn)，若AB的中點(diǎn)坐標(biāo)為(－2,1)，求直線l的方程．【解析】(1)因?yàn)殡x心率e＝eq\f(c,a)＝eq\f(\r(2),2)，所以a＝eq\r(2)c，因?yàn)閍2＝b2＋c2，所以b＝c.因?yàn)樗倪呅蜯F1NF2的面積為32，所以2bc＝32，所以b＝c＝4，a＝4eq\r(2)，故橢圓C的標(biāo)準(zhǔn)方程為eq\f(x2,32)＋eq\f(y2,16)＝1.(2)由題意得，直線l的斜率存在．設(shè)A(x1，y1)，B(x2，y2)，則eq\b\lc\{\rc\(\a\vs4\al\co1(\f(x\o\al(2,1),32)＋\f(y\o\al(2,1),16)＝1，,\f(x\o\al(2,2),32)＋\f(y\o\al(2,2),16)＝1，))兩式相減得eq\f(x\o\al(2,1)－x\o\al(2,2),32)＋eq\f(y\o\al(2,1)－y\o\al(2,2),16)＝0，所以eq\f(y1－y2,x1－x2)＝－eq\f(1,2)·eq\f(x1＋x2,y1＋y2).因?yàn)锳B的中點(diǎn)坐標(biāo)為(－2,1)在橢圓內(nèi)部，所以eq\f(y1－y2,x1－x2)＝1，所以直線l的斜率為1，故直線l的方程為y－1＝x＋2，即x－y＋3＝0.18．（2023·重慶·統(tǒng)考三模）已知橢圓SKIPIF1<0的上、下頂點(diǎn)分別為SKIPIF1<0，左頂點(diǎn)為SKIPIF1<0，SKIPIF1<0是面積為SKIPIF1<0的正三角形．(1)求橢圓SKIPIF1<0的方程；(2)過(guò)橢圓SKIPIF1<0外一點(diǎn)SKIPIF1<0的直線交橢圓SKIPIF1<0于SKIPIF1<0兩點(diǎn)，已知點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0軸對(duì)稱，點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0軸對(duì)稱，直線SKIPIF1<0與SKIPIF1<0交于點(diǎn)SKIPIF1<0，若SKIPIF1<0是鈍角，求SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）SKIPIF1<0是面積為SKIPIF1<0的正三角形，SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0橢圓SKIPIF1<0的方程為：SKIPIF1<0.（2）設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0直線SKIPIF1<0方程為：SKIPIF1<0，即SKIPIF1<0；由對(duì)稱性可知：點(diǎn)SKIPIF1<0在SKIPIF1<0軸上，則令SKIPIF1<0，解得：SKIPIF1<0，設(shè)直線SKIPIF1<0，由SKIPIF1<0得：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為鈍角，SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.19.（2023·安徽·校聯(lián)考三模）如圖，橢圓SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，點(diǎn)A，B，C分別為橢圓SKIPIF1<0的左、右頂點(diǎn)和上頂點(diǎn)，O為坐標(biāo)原點(diǎn)，過(guò)點(diǎn)SKIPIF1<0的直線l交橢圓SKIPIF1<0于E，F(xiàn)兩點(diǎn)，線段SKIPIF1<0的中點(diǎn)為SKIPIF1<0．點(diǎn)P是SKIPIF1<0上在第一象限內(nèi)的動(dòng)點(diǎn)，直線AP與直線BC相交于點(diǎn)Q，直線CP與x軸相交于點(diǎn)M．(1)求橢圓SKIPIF1<0的方程；(2)設(shè)SKIPIF1<0的面積為SKIPIF1<0，SKIPIF1<0的面積為SKIPIF1<0，求SKIPIF1<0的值．【答案】(1)SKIPIF1<0(2)16【詳解】（1）因?yàn)榫€段SKIPIF1<0的中點(diǎn)為SKIPIF1<0在y軸上，O為SKIPIF1<0的中點(diǎn)，所以SKIPIF1<0軸，即SKIPIF1<0軸，設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，代入橢圓SKIPIF1<0的方程得，SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以橢圓SKIPIF1<0的方程為SKIPIF1<0．（2）由題意可得SKIPIF1<0，SKIPIF1<0，所以直線BC的方程的截距式為SKIPIF1<0，即為SKIPIF1<0．設(shè)直線AP的斜率為k，點(diǎn)P的坐標(biāo)為SKIPIF1<0，則AP的方程為SKIPIF1<0，聯(lián)立SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0．所以SKIPIF1<0SKIPIF1<0．直線CP的方程為SKIPIF1<0，設(shè)點(diǎn)M，Q的坐標(biāo)分別為SKIPIF1<0，SKIPIF1<0，在SKIPIF1<0中，令SKIPIF1<0得SKIPIF1<0．解SKIPIF1<0得SKIPIF1<0．所以SKIPIF1<0．20．（2023·安徽蚌埠·統(tǒng)考三模）已知SKIPIF1<0，SKIPIF1<0是雙曲線SKIPIF1<0的左?右頂點(diǎn)，SKIPIF1<0為雙曲線上與SKIPIF1<0，SKIPIF1<0不重合的點(diǎn).(1)設(shè)直線SKIPIF1<0，SKIPIF1<0的斜率分別為SKIPIF1<0，SKIPIF1<0，求證：SKIPIF1<0是定值；(2)設(shè)直線SKIPIF1<0與直線SKIPIF1<0交于點(diǎn)SKIPIF1<0，SKIPIF1<0與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0滿足SKIPIF1<0，直線SKIPIF1<0與雙曲線SKIPIF1<0交于點(diǎn)SKIPIF1<0（與SKIPIF1<0，SKIPIF1<0，SKIPIF1<0不重合）.判斷直線SKIPIF1<0是否過(guò)定點(diǎn)，若直線SKIPIF1<0過(guò)定點(diǎn)，求出該定點(diǎn)坐標(biāo)；若直線SKIPIF1<0不過(guò)定點(diǎn)，請(qǐng)說(shuō)明理由.【答案】(1)證明見(jiàn)解析(2)直線MN過(guò)定點(diǎn)SKIPIF1<0【詳解】（1）設(shè)SKIPIF1<0，由題意SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0（2）設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，BN的斜率為SKIPIF1<0，由SKIPIF1<0知：SKIPIF1<0SKIPIF1<0，由(1)知：SKIPIF1<0所以SKIPIF1<0設(shè)MN：SKIPIF1<0，與雙曲線SKIPIF1<0聯(lián)立，得：SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0﹐則SKIPIF1<0整理得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍），故直線MN過(guò)定點(diǎn)SKIPIF1<0.21．（2023·山西運(yùn)城·統(tǒng)考三模）已知拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，SKIPIF1<0分別為SKIPIF1<0上兩個(gè)不同的動(dòng)點(diǎn)，SKIPIF1<0為坐標(biāo)原點(diǎn)，當(dāng)SKIPIF1<0為等邊三角形時(shí)，SKIPIF1<0.(1)求SKIPIF1<0的標(biāo)準(zhǔn)方程；(2)拋物線SKIPIF1<0在第一象限的部分是否存在點(diǎn)SKIPIF1<0，使得點(diǎn)SKIPIF1<0滿足SKIPIF1<0，且點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離為2？若存在，求出點(diǎn)SKIPIF1<0的坐標(biāo)及直線SKIPIF1<0的方程；若不存在，請(qǐng)說(shuō)明理由.【答案】(1)SKIPIF1<0(2)存在，點(diǎn)SKIPIF1<0，直線SKIPIF1<0的方程為SKIPIF1<0.【詳解】（1）由對(duì)稱性可知當(dāng)SKIPIF1<0為等邊三角形時(shí)，SKIPIF1<0兩點(diǎn)關(guān)于SKIPIF1<0軸對(duì)稱，當(dāng)SKIPIF1<0為等邊三角形時(shí)，SKIPIF1<0的高為SKIPIF1<0，由題意知點(diǎn)SKIPIF1<0在SKIPIF1<0上，代入SKIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的標(biāo)準(zhǔn)方程為SKIPIF1<0.（2）由（1）知SKIPIF1<0，根據(jù)題意可知直線SKIPIF1<0的斜率不為0，設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，聯(lián)立SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，又點(diǎn)SKIPIF1<0在SKIPIF1<0上，所以SKIPIF1<0，即SKIPIF1<0，①所以SKIPIF1<0，解得SKIPIF1<0，又點(diǎn)SKIPIF1<0在第一象限，所以SKIPIF1<0，所以SKIPIF1<0.又點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0，②聯(lián)立①②解得SKIPIF1<0，或SKIPIF1<0（舍去），或SKIPIF1<0（舍去）.此時(shí)點(diǎn)SKIPIF1<0，直線SKIPIF1<0的方程為SKIPIF1<0.22.（2023·山東淄博·統(tǒng)考二模）“工藝折紙”是一種把紙張折成各種不