新高考數(shù)學(xué)二輪復(fù)習(xí)培優(yōu)專題訓(xùn)練專題17 圓錐曲線中的雙曲線與拋物線問題（解析版）

專題17圓錐曲線中的雙曲線與拋物線問題1、（2023年全國甲卷數(shù)學(xué)（文）（理））已知雙曲線SKIPIF1<0的離心率為SKIPIF1<0，其中一條漸近線與圓SKIPIF1<0交于A，B兩點(diǎn)，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】由SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，所以雙曲線的一條漸近線不妨取SKIPIF1<0，則圓心SKIPIF1<0到漸近線的距離SKIPIF1<0，所以弦長(zhǎng)SKIPIF1<0.故選：D2、（2023年全國乙卷數(shù)學(xué)（文）(理)）設(shè)A，B為雙曲線SKIPIF1<0上兩點(diǎn)，下列四個(gè)點(diǎn)中，可為線段AB中點(diǎn)的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0的中點(diǎn)SKIPIF1<0，可得SKIPIF1<0，因?yàn)镾KIPIF1<0在雙曲線上，則SKIPIF1<0，兩式相減得SKIPIF1<0，所以SKIPIF1<0.對(duì)于選項(xiàng)A：可得SKIPIF1<0，則SKIPIF1<0，聯(lián)立方程SKIPIF1<0，消去y得SKIPIF1<0，此時(shí)SKIPIF1<0，所以直線AB與雙曲線沒有交點(diǎn)，故A錯(cuò)誤；對(duì)于選項(xiàng)B：可得SKIPIF1<0，則SKIPIF1<0，聯(lián)立方程SKIPIF1<0，消去y得SKIPIF1<0，此時(shí)SKIPIF1<0，所以直線AB與雙曲線沒有交點(diǎn)，故B錯(cuò)誤；對(duì)于選項(xiàng)C：可得SKIPIF1<0，則SKIPIF1<0由雙曲線方程可得SKIPIF1<0，則SKIPIF1<0為雙曲線的漸近線，所以直線AB與雙曲線沒有交點(diǎn)，故C錯(cuò)誤；對(duì)于選項(xiàng)D：SKIPIF1<0，則SKIPIF1<0，聯(lián)立方程SKIPIF1<0，消去y得SKIPIF1<0，此時(shí)SKIPIF1<0，故直線AB與雙曲線有交兩個(gè)交點(diǎn)，故D正確；故選：D.3、

【2022年全國乙卷】設(shè)F為拋物線C:y2=4x的焦點(diǎn)，點(diǎn)A在C上，點(diǎn)B(3,0)，若AFA．2 B．22 C．3 D．【答案】B【解析】由題意得，F(xiàn)1,0，則AF即點(diǎn)A到準(zhǔn)線x=?1的距離為2，所以點(diǎn)A的橫坐標(biāo)為?1+2=1，不妨設(shè)點(diǎn)A在x軸上方，代入得，A1,2所以AB=故選：B

4、【2022年全國乙卷】雙曲線C的兩個(gè)焦點(diǎn)為F1,F2，以C的實(shí)軸為直徑的圓記為D，過F1作D的切線與C的兩支交于M，NA．52 B．32 C．13【答案】C【解析】解：依題意不妨設(shè)雙曲線焦點(diǎn)在x軸，設(shè)過F1作圓D的切線切點(diǎn)為G所以O(shè)G⊥NF1，因?yàn)閏os∠所以O(shè)G=a，OF1=c，GF由cos∠F1NF2=35在△F2=sin由正弦定理得2csin所以NF1又NF所以2b=3a，即ba所以雙曲線的離心率e=故選：C5、（2023年新課標(biāo)全國Ⅱ卷）（多選題）設(shè)O為坐標(biāo)原點(diǎn)，直線SKIPIF1<0過拋物線SKIPIF1<0的焦點(diǎn)，且與C交于M，N兩點(diǎn)，l為C的準(zhǔn)線，則（

）．A．SKIPIF1<0 B．SKIPIF1<0C．以MN為直徑的圓與l相切 D．SKIPIF1<0為等腰三角形【答案】AC【詳解】A選項(xiàng)：直線SKIPIF1<0過點(diǎn)SKIPIF1<0，所以拋物線SKIPIF1<0的焦點(diǎn)SKIPIF1<0，所以SKIPIF1<0，則A選項(xiàng)正確，且拋物線SKIPIF1<0的方程為SKIPIF1<0.B選項(xiàng)：設(shè)SKIPIF1<0，由SKIPIF1<0消去SKIPIF1<0并化簡(jiǎn)得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，B選項(xiàng)錯(cuò)誤.C選項(xiàng)：設(shè)SKIPIF1<0的中點(diǎn)為SKIPIF1<0，SKIPIF1<0到直線SKIPIF1<0的距離分別為SKIPIF1<0，因?yàn)镾KIPIF1<0，即SKIPIF1<0到直線SKIPIF1<0的距離等于SKIPIF1<0的一半，所以以SKIPIF1<0為直徑的圓與直線SKIPIF1<0相切，C選項(xiàng)正確.D選項(xiàng)：直線SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0，所以三角形SKIPIF1<0的面積為SKIPIF1<0，由上述分析可知SKIPIF1<0，所以SKIPIF1<0，所以三角形SKIPIF1<0不是等腰三角形，D選項(xiàng)錯(cuò)誤.故選：AC.6、【2022年新高考1卷】（多選題）已知O為坐標(biāo)原點(diǎn)，點(diǎn)A(1,1)在拋物線C:x2=2py(p>0)上，過點(diǎn)B(0,?1)的直線交C于PA．C的準(zhǔn)線為y=?1 B．直線AB與C相切C．|OP|?|OQ|>|OA2 D．【答案】BCD【解析】將點(diǎn)A的代入拋物線方程得1=2p，所以拋物線方程為x2=y，故準(zhǔn)線方程為kAB=1?(?1)1?0=2聯(lián)立y=2x?1x2=y，可得x設(shè)過B的直線為l，若直線l與y軸重合，則直線l與拋物線C只有一個(gè)交點(diǎn)，所以，直線l的斜率存在，設(shè)其方程為y=kx?1，P(x聯(lián)立y=kx?1x2=y所以Δ=k2?4>0x1+又|OP|=x12所以|OP|?|OQ|=y因?yàn)閨BP|=1+k2所以|BP|?|BQ|=(1+k2)|故選：BCD

7、【2022年新高考2卷】（多選題）已知O為坐標(biāo)原點(diǎn)，過拋物線C:y2=2px(p>0)焦點(diǎn)F的直線與C交于A，B兩點(diǎn)，其中A在第一象限，點(diǎn)M(p,0)A．直線AB的斜率為26 B．C．|AB|>4|OF| D．∠OAM+∠OBM<180°【答案】ACD【解析】對(duì)于A，易得F(p2,0)，由AF=AM可得點(diǎn)A在FM代入拋物線可得y2=2p?3p4=32對(duì)于B，由斜率為26可得直線AB的方程為x=12設(shè)B(x1,y1)，則62p+y則OB=對(duì)于C，由拋物線定義知：AB=對(duì)于D，OA?OB=(又MA?MB=(?又∠AOB+∠AMB+∠OAM+∠OBM=360°，則故選：ACD.8、（2023年新課標(biāo)全國Ⅰ卷）已知雙曲線SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0．點(diǎn)SKIPIF1<0在SKIPIF1<0上，點(diǎn)SKIPIF1<0在SKIPIF1<0軸上，SKIPIF1<0，則SKIPIF1<0的離心率為________．【答案】SKIPIF1<0/SKIPIF1<0【詳解】方法一：依題意，設(shè)SKIPIF1<0，則SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0或SKIPIF1<0（舍去），所以SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，所以在SKIPIF1<0中，SKIPIF1<0，整理得SKIPIF1<0，故SKIPIF1<0.方法二:依題意，得SKIPIF1<0，令SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，則SKIPIF1<0，又點(diǎn)SKIPIF1<0在SKIPIF1<0上，則SKIPIF1<0，整理得SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，整理得SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0（舍去），故SKIPIF1<0.故答案為：SKIPIF1<0.9、【2022年全國甲卷】記雙曲線C:x2a2?y2b2【答案】2（滿足1<e≤5【解析】C:x2a2?結(jié)合漸近線的特點(diǎn)，只需0<ba≤2可滿足條件“直線y=2x與C無公共點(diǎn)”所以e=c又因?yàn)閑>1，所以1<e≤5故答案為：2（滿足1<e≤510、（2023年全國乙卷數(shù)學(xué)（文）(理)）已知點(diǎn)SKIPIF1<0在拋物線C：SKIPIF1<0上，則A到C的準(zhǔn)線的距離為______.【答案】SKIPIF1<0【詳解】由題意可得：SKIPIF1<0，則SKIPIF1<0，拋物線的方程為SKIPIF1<0，準(zhǔn)線方程為SKIPIF1<0，點(diǎn)SKIPIF1<0到SKIPIF1<0的準(zhǔn)線的距離為SKIPIF1<0.故答案為：SKIPIF1<0.題組一、雙曲線的離心率1-1、（2023·江蘇泰州·泰州中學(xué)?？家荒＃┰谄矫嬷苯亲鴺?biāo)系SKIPIF1<0中，SKIPIF1<0分別是雙曲線C：SKIPIF1<0的左，右焦點(diǎn)，過SKIPIF1<0的直線SKIPIF1<0與雙曲線的左，右兩支分別交于點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0在SKIPIF1<0軸上，滿足SKIPIF1<0，且SKIPIF1<0經(jīng)過SKIPIF1<0的內(nèi)切圓圓心，則雙曲線SKIPIF1<0的離心率為（

）A．SKIPIF1<0 B．2 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)雙曲線的定義先推出SKIPIF1<0為正三角形，然后根據(jù)余弦定理解決.【詳解】SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0經(jīng)過SKIPIF1<0內(nèi)切圓圓心，∴SKIPIF1<0為SKIPIF1<0的角平分線，∴SKIPIF1<0．∴SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0，于是SKIPIF1<0，∴SKIPIF1<0為正三角形，SKIPIF1<0．SKIPIF1<0中，由余弦定理，SKIPIF1<0∴SKIPIF1<0.故選：C.1-2、（2023·黑龍江·黑龍江實(shí)驗(yàn)中學(xué)?？家荒＃┮阎p曲線SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0，一條漸近線為l，過點(diǎn)SKIPIF1<0且與l平行的直線交雙曲線C于點(diǎn)M，若SKIPIF1<0，則雙曲線C的離心率為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．3【答案】B【分析】根據(jù)雙曲線的定義，結(jié)合余弦定理、同角的三角函數(shù)關(guān)系式進(jìn)行求解即可.【詳解】根據(jù)雙曲線的對(duì)稱性，不妨設(shè)一條漸近線l的方程為SKIPIF1<0，因此直線SKIPIF1<0的傾斜角SKIPIF1<0的正切值為SKIPIF1<0，即SKIPIF1<0，所以有SKIPIF1<0，設(shè)SKIPIF1<0，由雙曲線定義可知：SKIPIF1<0SKIPIF1<0，由余弦定理可知：SKIPIF1<0，故選：B.1-3、（2023·吉林通化·梅河口市第五中學(xué)校考一模）已知雙曲線C：SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）的左、右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，若在C上存在點(diǎn)P（不是頂點(diǎn)），使得SKIPIF1<0，則C的離心率的取值范圍為______．【答案】SKIPIF1<0【分析】SKIPIF1<0與SKIPIF1<0軸交點(diǎn)SKIPIF1<0，連接SKIPIF1<0，由雙曲線的定義和對(duì)稱性，結(jié)合已知條件得SKIPIF1<0，有SKIPIF1<0且SKIPIF1<0，可求離心率的取值范圍.【詳解】設(shè)SKIPIF1<0與SKIPIF1<0軸交點(diǎn)SKIPIF1<0，連接SKIPIF1<0，由對(duì)稱性可知，SKIPIF1<0，如圖所示，又∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0．又∵SKIPIF1<0，∴SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，由SKIPIF1<0，且三角形的內(nèi)角和為SKIPIF1<0，SKIPIF1<0SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0綜上，SKIPIF1<0．故答案為：SKIPIF1<0．題組二、雙曲線與拋物線的性質(zhì)2-1、（2023·江蘇南通·統(tǒng)考模擬預(yù)測(cè)）（多選題）已知雙曲線SKIPIF1<0的右頂點(diǎn)為A，右焦點(diǎn)為F，雙曲線上一點(diǎn)P滿足PA=2，則PF的長(zhǎng)度可能為（

）A．2 B．3 C．4 D．5【答案】AB【分析】設(shè)SKIPIF1<0，根據(jù)點(diǎn)P在雙曲線上且PA=2，則可求得SKIPIF1<0的值，從而可求得SKIPIF1<0的值，進(jìn)而可求得PF的長(zhǎng)度．【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0．故選：AB．2-2、（2023·江蘇南京·?？家荒＃┮阎獧E圓SKIPIF1<0的兩個(gè)焦點(diǎn)為SKIPIF1<0和SKIPIF1<0，直線l過點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0關(guān)于l的對(duì)稱點(diǎn)A在C上，且SKIPIF1<0，則C的方程為__________．【答案】SKIPIF1<0【分析】根據(jù)向量的線性運(yùn)算和數(shù)量積的性質(zhì)化簡(jiǎn)SKIPIF1<0，由條件結(jié)合橢圓的定義可求SKIPIF1<0，由SKIPIF1<0求SKIPIF1<0，可得橢圓方程.【詳解】因?yàn)锳與SKIPIF1<0關(guān)于直線l對(duì)稱，所以直線l為SKIPIF1<0的垂直平分線，又SKIPIF1<0，SKIPIF1<0所以SKIPIF1<0，由橢圓的定義可得SKIPIF1<0，設(shè)直線l與SKIPIF1<0交于點(diǎn)M，則M為SKIPIF1<0的中點(diǎn)，且SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，解得SKIPIF1<0或1（舍去），所以SKIPIF1<0，SKIPIF1<0，則C的方程為：SKIPIF1<0．故答案為：SKIPIF1<0．2-3、（2023·山西·統(tǒng)考一模）已知拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，點(diǎn)SKIPIF1<0，SKIPIF1<0為拋物線上一動(dòng)點(diǎn)，則SKIPIF1<0周長(zhǎng)的最小值為______.【答案】SKIPIF1<0【分析】過SKIPIF1<0作準(zhǔn)線的垂線，垂足為SKIPIF1<0，過SKIPIF1<0作準(zhǔn)線的垂線，垂足為SKIPIF1<0，進(jìn)而結(jié)合拋物線的定義求解即可.【詳解】解：由題知SKIPIF1<0，準(zhǔn)線方程為SKIPIF1<0.如圖，過SKIPIF1<0作準(zhǔn)線的垂線，垂足為SKIPIF1<0，過SKIPIF1<0作準(zhǔn)線的垂線，垂足為SKIPIF1<0，所以SKIPIF1<0周長(zhǎng)SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0為SKIPIF1<0與拋物線的交點(diǎn)SKIPIF1<0時(shí)等號(hào)成立.故答案為：SKIPIF1<0.2-4、（2023·江蘇南通·統(tǒng)考模擬預(yù)測(cè)）已知拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，點(diǎn)SKIPIF1<0在拋物線上，且SKIPIF1<0，若SKIPIF1<0的面積為SKIPIF1<0，則SKIPIF1<0（

）A．2 B．4 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】根據(jù)拋物線定義求得SKIPIF1<0點(diǎn)橫坐標(biāo)，代入拋物線方程得縱坐標(biāo)，再利用三角形面積公式即可得SKIPIF1<0的值.【詳解】拋物線的焦點(diǎn)為SKIPIF1<0，點(diǎn)SKIPIF1<0在拋物線上，由拋物線的定義可得SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍）.故選：B.題組三、拋物線、雙曲線、橢圓的綜合應(yīng)用3-1、（2023·江蘇南通·統(tǒng)考一模）已知拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，以該拋物線上三點(diǎn)SKIPIF1<0為切點(diǎn)的切線分別是SKIPIF1<0，直線SKIPIF1<0相交于點(diǎn)SKIPIF1<0與SKIPIF1<0分別相交于點(diǎn)SKIPIF1<0.記SKIPIF1<0的橫坐標(biāo)分別為SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BCD【分析】利用導(dǎo)函數(shù)和斜率的關(guān)系表示出切線方程可求出SKIPIF1<0的坐標(biāo)可判斷A，根據(jù)向量數(shù)量積的坐標(biāo)運(yùn)算判斷B,并根據(jù)兩點(diǎn)間的距離公式運(yùn)算求解即可判斷C,D.【詳解】設(shè)SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，同理SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，也即SKIPIF1<0，B正確；SKIPIF1<0SKIPIF1<0SKIPIF1<0不一定為SKIPIF1<0A錯(cuò)誤；SKIPIF1<0SKIPIF1<0SKIPIF1<0正確；SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0正確，故選：BCD.3-2、（2023·黑龍江·黑龍江實(shí)驗(yàn)中學(xué)?？家荒＃┮阎狿為拋物線SKIPIF1<0上的動(dòng)點(diǎn)，SKIPIF1<0為坐標(biāo)原點(diǎn)，SKIPIF1<0在拋物線C上，過拋物線C的焦點(diǎn)F的直線l與拋物線C交于A，B兩點(diǎn)，SKIPIF1<0，則（

）A．SKIPIF1<0的最小值為4B．若線段AB的中點(diǎn)為M，則弦長(zhǎng)AB的長(zhǎng)度為8C．若線段AB的中點(diǎn)為M，則三角形OAB的面積為SKIPIF1<0D．過點(diǎn)SKIPIF1<0作兩條直線與拋物線C分別交于點(diǎn)G，H，且滿足EF平分SKIPIF1<0，則直線GH的斜率為定值【答案】ABD【分析】先求出拋物線的方程SKIPIF1<0，利用拋物線的定義轉(zhuǎn)化即可求出最小值可判斷A；由直線與拋物線相交的弦長(zhǎng)公式判斷B；由點(diǎn)到直線的距離公式可求三角形OAB的面積判斷C；設(shè)SKIPIF1<0，SKIPIF1<0,將已知轉(zhuǎn)化為SKIPIF1<0結(jié)合兩點(diǎn)連線的斜率公式即可判斷直線GH的斜率是否為定值判斷D.【詳解】由SKIPIF1<0在拋物線C上，得SKIPIF1<0，拋物線C的方程為SKIPIF1<0，SKIPIF1<0．對(duì)于A，過點(diǎn)P作拋物線的準(zhǔn)線SKIPIF1<0的垂線PD，垂足為D，由拋物線的定義知SKIPIF1<0，即M，P，D三點(diǎn)共線時(shí)，SKIPIF1<0取得最小值，為SKIPIF1<0，故A正確．對(duì)于B，因?yàn)镾KIPIF1<0為AB的中點(diǎn)，點(diǎn)A，B的橫坐標(biāo)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，故B正確．對(duì)于C，由直線l過焦點(diǎn)SKIPIF1<0與SKIPIF1<0求得直線l的方程為SKIPIF1<0，則點(diǎn)O到直線l的距離SKIPIF1<0，則SKIPIF1<0，故C錯(cuò)誤；對(duì)于D，易知點(diǎn)SKIPIF1<0在拋物線上且SKIPIF1<0軸．設(shè)SKIPIF1<0，SKIPIF1<0．易知直線EG，EH的斜率存在，SKIPIF1<0，同理SKIPIF1<0．因?yàn)镋F平分SKIPIF1<0，SKIPIF1<0軸，所以SKIPIF1<0，即SKIPIF1<0，直線SKIPIF1<0，所以SKIPIF1<0，直線GH的斜率SKIPIF1<0為定值，故D正確．故選：ABD.3-3、（2023·江蘇蘇州·蘇州中學(xué)?？寄M預(yù)測(cè)）已知拋物線SKIPIF1<0：SKIPIF1<0，圓SKIPIF1<0：SKIPIF1<0，在拋物線SKIPIF1<0上任取一點(diǎn)SKIPIF1<0，向圓SKIPIF1<0作兩條切線SKIPIF1<0和SKIPIF1<0，切點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的取值范圍是______．【答案】SKIPIF1<0【分析】設(shè)點(diǎn)SKIPIF1<0，由已知關(guān)系，可用SKIPIF1<0點(diǎn)坐標(biāo)表示出SKIPIF1<0.在SKIPIF1<0，有SKIPIF1<0，進(jìn)而可推出SKIPIF1<0，根據(jù)SKIPIF1<0的范圍，即可得到結(jié)果.【詳解】由已知，SKIPIF1<0，SKIPIF1<0.如圖，設(shè)點(diǎn)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0SKIPIF1<0，在SKIPIF1<0中，有SKIPIF1<0SKIPIF1<0，易知SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，因?yàn)椋琒KIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最大值SKIPIF1<0，又SKIPIF1<0，所以，SKIPIF1<0.所以，SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<01、（2023·江蘇南通·統(tǒng)考一模）2022年神舟接力騰飛，中國空間站全面建成，我們的“太空之家”遨游蒼穹.太空中飛船與空間站的對(duì)接，需要經(jīng)過多次變軌.某飛船升空后的初始運(yùn)行軌道是以地球的中心為一個(gè)焦點(diǎn)的橢圓，其遠(yuǎn)地點(diǎn)（長(zhǎng)軸端點(diǎn)中離地面最遠(yuǎn)的點(diǎn)）距地面SKIPIF1<0，近地點(diǎn)（長(zhǎng)軸端點(diǎn)中離地面最近的點(diǎn)）距地面SKIPIF1<0，地球的半徑為SKIPIF1<0，則該橢圓的短軸長(zhǎng)為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)橢圓的遠(yuǎn)地點(diǎn)和近地點(diǎn)的距離可得SKIPIF1<0，進(jìn)而可求得SKIPIF1<0，求得b，可得答案.【詳解】由題意得SKIPIF1<0，故SKIPIF1<0，故選：D.2、（2022·山東青島·高三期末）已知坐標(biāo)原點(diǎn)為SKIPIF1<0，雙曲線SKIPIF1<0的右焦點(diǎn)為SKIPIF1<0，點(diǎn)SKIPIF1<0，若SKIPIF1<0，則雙曲線SKIPIF1<0的離心率為（）A．2 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】設(shè)雙曲線SKIPIF1<0的右頂點(diǎn)為SKIPIF1<0，又點(diǎn)SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0垂直平分線段SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0.故選：A3、（2022·湖北襄陽·高三期末）若雙曲線SKIPIF1<0的一條漸近線被圓SKIPIF1<0所截得的弦長(zhǎng)為2，則雙曲線的離心率為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】不妨設(shè)雙曲線SKIPIF1<0SKIPIF1<0的一條漸近線為SKIPIF1<0，圓SKIPIF1<0的圓心為SKIPIF1<0，半徑SKIPIF1<0，則圓心到漸近線的距離為SKIPIF1<0所以弦長(zhǎng)SKIPIF1<0，化簡(jiǎn)得：SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0.故選：B4、（2023·江蘇徐州·徐州市第七中學(xué)?？家荒＃ǘ噙x題）已知SKIPIF1<0，SKIPIF1<0是拋物線SKIPIF1<0：SKIPIF1<0上兩動(dòng)點(diǎn)，SKIPIF1<0為拋物線SKIPIF1<0的焦點(diǎn)，則（

）A．直線SKIPIF1<0過焦點(diǎn)SKIPIF1<0時(shí)，SKIPIF1<0最小值為2B．直線SKIPIF1<0過焦點(diǎn)SKIPIF1<0且傾斜角為60°時(shí)（點(diǎn)SKIPIF1<0在第一象限），SKIPIF1<0C．若SKIPIF1<0中點(diǎn)SKIPIF1<0的橫坐標(biāo)為3，則SKIPIF1<0最大值為8D．點(diǎn)SKIPIF1<0坐標(biāo)SKIPIF1<0，且直線SKIPIF1<0，SKIPIF1<0斜率之和為0，SKIPIF1<0與拋物線的另一交點(diǎn)為SKIPIF1<0，則直線SKIPIF1<0方程為：SKIPIF1<0【答案】CD【分析】對(duì)于AB項(xiàng)畫出函數(shù)圖像，把SKIPIF1<0用直線SKIPIF1<0的傾斜角表示，驗(yàn)證是否正確；對(duì)于C項(xiàng)，SKIPIF1<0可求解；對(duì)于D項(xiàng)根據(jù)點(diǎn)SKIPIF1<0可求出SKIPIF1<0，就能求出SKIPIF1<0所以求出直線SKIPIF1<0，SKIPIF1<0分別與拋物線聯(lián)立求出SKIPIF1<0點(diǎn)，就能求出SKIPIF1<0方程.【詳解】對(duì)于A項(xiàng)，過點(diǎn)SKIPIF1<0分別作準(zhǔn)線SKIPIF1<0的垂線，垂足分別為SKIPIF1<0，過點(diǎn)SKIPIF1<0分別作SKIPIF1<0軸的垂線，垂足分別為SKIPIF1<0，準(zhǔn)線與SKIPIF1<0軸的交點(diǎn)為SKIPIF1<0，設(shè)直線SKIPIF1<0的傾斜角為SKIPIF1<0，畫圖為：根據(jù)拋物線的定義：SKIPIF1<0，從圖可知SKIPIF1<0，SKIPIF1<0SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，所以SKIPIF1<0，同理SKIPIF1<0則SKIPIF1<0SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)SKIPIF1<0故SKIPIF1<0最小值為SKIPIF1<0，所以A不正確.對(duì)于B項(xiàng)，由A可知，SKIPIF1<0，SKIPIF1<0所以SKIPIF1<0，故B不正確.對(duì)于C項(xiàng)，SKIPIF1<0所以SKIPIF1<0最大值為8，故C正確.對(duì)于D項(xiàng)，由SKIPIF1<0，SKIPIF1<0，知SKIPIF1<0，所以SKIPIF1<0所以直線SKIPIF1<0的方程為SKIPIF1<0，直線SKIPIF1<0的方程為SKIPIF1<0聯(lián)立SKIPIF1<0解得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0聯(lián)立SKIPIF1<0解得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0所以直線SKIPIF1<0的方程為SKIPIF1<0即SKIPIF1<0，故D正確.故選：CD5、（2023·安徽·統(tǒng)考一模）（多選題）已知SKIPIF1<0為坐標(biāo)原點(diǎn)，點(diǎn)SKIPIF1<0，線段SKIPIF1<0的中點(diǎn)SKIPIF1<0在拋物線SKIPIF1<0上，連接SKIPIF1<0并延長(zhǎng)，與SKIPIF1<0交于點(diǎn)SKIPIF1<0，則（

）A．SKIPIF1<0的準(zhǔn)線方程為SKIPIF1<0 B．點(diǎn)SKIPIF1<0為線段SKIPIF1<0的中點(diǎn)C．直線SKIPIF1<0與SKIPIF1<0相切 D．SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0平行【答案】BCD【分析】將SKIPIF1<0代入拋物線得SKIPIF1<0，則得到其準(zhǔn)線方程，則可判斷A，聯(lián)立直線SKIPIF1<0的方程與拋物線方程即可得到SKIPIF1<0，即可判斷B，利用導(dǎo)數(shù)求出拋物線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程，令SKIPIF1<0，則可判斷C，再次利用導(dǎo)數(shù)求出拋物線在SKIPIF1<0處的切線斜率，則可判斷D.【詳解】對(duì)A，根據(jù)中點(diǎn)公式得SKIPIF1<0，將其代入SKIPIF1<0得SKIPIF1<0，則SKIPIF1<0，所以拋物線SKIPIF1<0的準(zhǔn)線方程為SKIPIF1<0，故A錯(cuò)誤，對(duì)B，SKIPIF1<0，則直線SKIPIF1<0的斜率為SKIPIF1<0，則直線SKIPIF1<0的方程為SKIPIF1<0，將其代入SKIPIF1<0