2024年高中數(shù)學(xué)新高二暑期培優(yōu)講義第15講 直線(xiàn)和圓錐曲線(xiàn)的位置關(guān)系（教師版）

第第頁(yè)第15講直線(xiàn)和圓錐曲線(xiàn)的位置關(guān)系【題型歸納目錄】題型一：直線(xiàn)與橢圓的位置關(guān)系題型二：橢圓的弦題型三：橢圓的綜合問(wèn)題題型四：直線(xiàn)與雙曲線(xiàn)的位置關(guān)系題型五：雙曲線(xiàn)的弦題型六：雙曲線(xiàn)的綜合問(wèn)題題型七：直線(xiàn)與拋物線(xiàn)的位置關(guān)系題型八：拋物線(xiàn)的弦題型九：拋物線(xiàn)的綜合問(wèn)題【知識(shí)點(diǎn)梳理】知識(shí)點(diǎn)一：直線(xiàn)與橢圓的位置關(guān)系平面內(nèi)點(diǎn)與橢圓的位置關(guān)系橢圓將平面分成三部分：橢圓上、橢圓內(nèi)、橢圓外，因此，平面上的點(diǎn)與橢圓的位置關(guān)系有三種，任給一點(diǎn)M(x，y)，若點(diǎn)M(x，y)在橢圓上，則有SKIPIF1<0SKIPIF1<0；若點(diǎn)M(x，y)在橢圓內(nèi)，則有SKIPIF1<0SKIPIF1<0；若點(diǎn)M(x，y)在橢圓外，則有SKIPIF1<0SKIPIF1<0.直線(xiàn)與橢圓的位置關(guān)系將直線(xiàn)的方程SKIPIF1<0與橢圓的方程SKIPIF1<0SKIPIF1<0聯(lián)立成方程組，消元轉(zhuǎn)化為關(guān)于x或y的一元二次方程，其判別式為Δ.①Δ＞0SKIPIF1<0直線(xiàn)和橢圓相交SKIPIF1<0直線(xiàn)和橢圓有兩個(gè)交點(diǎn)(或兩個(gè)公共點(diǎn))；②Δ＝0SKIPIF1<0直線(xiàn)和橢圓相切SKIPIF1<0直線(xiàn)和橢圓有一個(gè)切點(diǎn)(或一個(gè)公共點(diǎn))；③Δ＜0SKIPIF1<0直線(xiàn)和橢圓相離SKIPIF1<0直線(xiàn)和橢圓無(wú)公共點(diǎn)．直線(xiàn)與橢圓的相交弦設(shè)直線(xiàn)SKIPIF1<0交橢圓SKIPIF1<0SKIPIF1<0于點(diǎn)SKIPIF1<0兩點(diǎn)，則SKIPIF1<0=SKIPIF1<0=SKIPIF1<0同理可得SKIPIF1<0這里SKIPIF1<0SKIPIF1<0的求法通常使用韋達(dá)定理，需作以下變形：SKIPIF1<0SKIPIF1<0知識(shí)點(diǎn)三、直線(xiàn)與雙曲線(xiàn)的位置關(guān)系直線(xiàn)與雙曲線(xiàn)的位置關(guān)系將直線(xiàn)的方程SKIPIF1<0與雙曲線(xiàn)的方程SKIPIF1<0SKIPIF1<0聯(lián)立成方程組，消元轉(zhuǎn)化為關(guān)于x或y的一元二次方程，其判別式為Δ.SKIPIF1<0若SKIPIF1<0即SKIPIF1<0，直線(xiàn)與雙曲線(xiàn)漸近線(xiàn)平行，直線(xiàn)與雙曲線(xiàn)相交于一點(diǎn)；若SKIPIF1<0即SKIPIF1<0，①Δ＞0SKIPIF1<0直線(xiàn)和雙曲線(xiàn)相交SKIPIF1<0直線(xiàn)和雙曲線(xiàn)相交，有兩個(gè)交點(diǎn)；②Δ＝0SKIPIF1<0直線(xiàn)和雙曲線(xiàn)相切SKIPIF1<0直線(xiàn)和雙曲線(xiàn)相切，有一個(gè)公共點(diǎn)；③Δ＜0SKIPIF1<0直線(xiàn)和雙曲線(xiàn)相離SKIPIF1<0直線(xiàn)和雙曲線(xiàn)相離，無(wú)公共點(diǎn)．直線(xiàn)與雙曲線(xiàn)的相交弦設(shè)直線(xiàn)SKIPIF1<0交雙曲線(xiàn)SKIPIF1<0SKIPIF1<0于點(diǎn)SKIPIF1<0兩點(diǎn)，則SKIPIF1<0=SKIPIF1<0=SKIPIF1<0同理可得SKIPIF1<0這里SKIPIF1<0SKIPIF1<0的求法通常使用韋達(dá)定理，需作以下變形：SKIPIF1<0SKIPIF1<0雙曲線(xiàn)的中點(diǎn)弦問(wèn)題遇到中點(diǎn)弦問(wèn)題常用“韋達(dá)定理”或“點(diǎn)差法”求解.在雙曲線(xiàn)SKIPIF1<0SKIPIF1<0中，以SKIPIF1<0為中點(diǎn)的弦所在直線(xiàn)的斜率SKIPIF1<0；涉及弦長(zhǎng)的中點(diǎn)問(wèn)題，常用“點(diǎn)差法”設(shè)而不求，將弦所在直線(xiàn)的斜率、弦的中點(diǎn)坐標(biāo)聯(lián)系起來(lái)相互轉(zhuǎn)化，同時(shí)還應(yīng)充分挖掘題目的隱含條件，尋找量與量間的關(guān)系靈活轉(zhuǎn)化，往往就能事半功倍.解題的主要規(guī)律可以概括為“聯(lián)立方程求交點(diǎn)，韋達(dá)定理求弦長(zhǎng)，根的分布找范圍，曲線(xiàn)定義不能忘”.知識(shí)點(diǎn)四、直線(xiàn)與拋物線(xiàn)的位置關(guān)系直線(xiàn)與拋物線(xiàn)的位置關(guān)系將直線(xiàn)的方程SKIPIF1<0與拋物線(xiàn)的方程y2=2px(p＞0)聯(lián)立成方程組，消元轉(zhuǎn)化為關(guān)于x或y的一元二次方程，其判別式為Δ.SKIPIF1<0若SKIPIF1<0，直線(xiàn)與拋物線(xiàn)的對(duì)稱(chēng)軸平行或重合，直線(xiàn)與拋物線(xiàn)相交于一點(diǎn)；若SKIPIF1<0①Δ＞0SKIPIF1<0直線(xiàn)和拋物線(xiàn)相交，有兩個(gè)交點(diǎn)；②Δ＝0SKIPIF1<0直線(xiàn)和拋物線(xiàn)相切，有一個(gè)公共點(diǎn)；③Δ＜0SKIPIF1<0直線(xiàn)和拋物線(xiàn)相離，無(wú)公共點(diǎn)．直線(xiàn)與拋物線(xiàn)的相交弦設(shè)直線(xiàn)SKIPIF1<0交拋物線(xiàn)SKIPIF1<0SKIPIF1<0于點(diǎn)SKIPIF1<0兩點(diǎn)，則SKIPIF1<0=SKIPIF1<0=SKIPIF1<0同理可得SKIPIF1<0這里SKIPIF1<0SKIPIF1<0的求法通常使用韋達(dá)定理，需作以下變形：SKIPIF1<0SKIPIF1<0拋物線(xiàn)的焦點(diǎn)弦問(wèn)題已知過(guò)拋物線(xiàn)SKIPIF1<0的焦點(diǎn)F的直線(xiàn)交拋物線(xiàn)于A(yíng)、B兩點(diǎn)。設(shè)A(x1，y1)，B(x2，y2)，則：焦點(diǎn)弦長(zhǎng)SKIPIF1<0②SKIPIF1<0③SKIPIF1<0，其中|AF|叫做焦半徑，SKIPIF1<0④焦點(diǎn)弦長(zhǎng)最小值為2p。根據(jù)SKIPIF1<0時(shí)，即AB垂直于x軸時(shí)，弦AB的長(zhǎng)最短，最短值為2p?！镜淅}】題型一：直線(xiàn)與橢圓的位置關(guān)系例1．若直線(xiàn)SKIPIF1<0與橢圓SKIPIF1<0有且只有一公共點(diǎn)，那么SKIPIF1<0的值為(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因?yàn)榉匠蘏KIPIF1<0表示的曲線(xiàn)為橢圓，則SKIPIF1<0，將直線(xiàn)SKIPIF1<0的方程與橢圓的方程聯(lián)立，SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0.故選：C.例2．已知橢圓SKIPIF1<0，直線(xiàn)SKIPIF1<0，則直線(xiàn)l與橢圓C的位置關(guān)系為(

)A．相交 B．相切 C．相離 D．不確定【答案】A【解析】對(duì)于直線(xiàn)SKIPIF1<0，整理得SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，故直線(xiàn)SKIPIF1<0過(guò)定點(diǎn)SKIPIF1<0.∵SKIPIF1<0，則點(diǎn)SKIPIF1<0在橢圓C的內(nèi)部，所以直線(xiàn)l與橢圓C相交.故選：A.題型二：橢圓的弦例3．已知橢圓SKIPIF1<0的左焦點(diǎn)為SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0且傾斜角為SKIPIF1<0的直線(xiàn)SKIPIF1<0與橢圓SKIPIF1<0相交于SKIPIF1<0兩點(diǎn)，則SKIPIF1<0__________．【答案】SKIPIF1<0【解析】已知橢圓SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，所以橢圓的左焦點(diǎn)為SKIPIF1<0,因?yàn)橹本€(xiàn)SKIPIF1<0傾斜角為SKIPIF1<0，所以直線(xiàn)SKIPIF1<0的斜率SKIPIF1<0，則直線(xiàn)SKIPIF1<0的方程為SKIPIF1<0.聯(lián)立SKIPIF1<0，消去SKIPIF1<0，整理得SKIPIF1<0，解得SKIPIF1<0．SKIPIF1<0．故答案為：SKIPIF1<0.例4．橢圓SKIPIF1<0的焦點(diǎn)為SKIPIF1<0、SKIPIF1<0，過(guò)O作直線(xiàn)交橢圓于A(yíng)、B兩點(diǎn)，若SKIPIF1<0的面積為20，則直線(xiàn)AB的方程為_(kāi)_____．【答案】SKIPIF1<0或SKIPIF1<0【解析】由直線(xiàn)AB關(guān)于原點(diǎn)對(duì)稱(chēng)以及橢圓關(guān)于原點(diǎn)對(duì)稱(chēng)可知，SKIPIF1<0，從而SKIPIF1<0．過(guò)點(diǎn)A作AH垂直于x軸，垂足為H，則SKIPIF1<0，即點(diǎn)A的縱坐標(biāo)為SKIPIF1<0，代入橢圓方程解得A的橫坐標(biāo)為SKIPIF1<0，即點(diǎn)A的坐標(biāo)為SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0．因此直線(xiàn)AB的方程為SKIPIF1<0或SKIPIF1<0．故答案為：SKIPIF1<0或SKIPIF1<0題型三：橢圓的綜合問(wèn)題例5．已知橢圓SKIPIF1<0：SKIPIF1<0(SKIPIF1<0)上任意一點(diǎn)SKIPIF1<0到兩個(gè)焦點(diǎn)的距離之和為SKIPIF1<0，且離心率為SKIPIF1<0．(1)求橢圓SKIPIF1<0的標(biāo)準(zhǔn)方程；(2)過(guò)點(diǎn)SKIPIF1<0作直線(xiàn)SKIPIF1<0交橢圓于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，點(diǎn)SKIPIF1<0為線(xiàn)段SKIPIF1<0的中點(diǎn)，求直線(xiàn)SKIPIF1<0的方程．【解析】(1)由橢圓的定義知，SKIPIF1<0，∴SKIPIF1<0，又∵橢圓的離心率SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴橢圓SKIPIF1<0的標(biāo)準(zhǔn)方程為SKIPIF1<0.(2)∵SKIPIF1<0為橢圓SKIPIF1<0內(nèi)一點(diǎn)，∴直線(xiàn)SKIPIF1<0與橢圓必交于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，設(shè)SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，不合題意，故SKIPIF1<0，∵SKIPIF1<0為線(xiàn)段SKIPIF1<0的中點(diǎn)，∴SKIPIF1<0，∴SKIPIF1<0，又∵SKIPIF1<0，SKIPIF1<0均在橢圓上，∴SKIPIF1<0，兩式相減，得SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，∴直線(xiàn)SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0．題型四：直線(xiàn)與雙曲線(xiàn)的位置關(guān)系例6．直線(xiàn)SKIPIF1<0與雙曲線(xiàn)SKIPIF1<0相交，有且只有1個(gè)交點(diǎn)，則雙曲線(xiàn)SKIPIF1<0的離心率為(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因?yàn)橹本€(xiàn)SKIPIF1<0與雙曲線(xiàn)SKIPIF1<0：SKIPIF1<0相交，且有且僅有1個(gè)交點(diǎn),所以直線(xiàn)SKIPIF1<0與雙曲線(xiàn)SKIPIF1<0：SKIPIF1<0的漸近線(xiàn)SKIPIF1<0平行，故SKIPIF1<0，則雙曲線(xiàn)SKIPIF1<0的離心率SKIPIF1<0.故選：A例7．若直線(xiàn)SKIPIF1<0與曲線(xiàn)SKIPIF1<0有且只有一個(gè)交點(diǎn)，則滿(mǎn)足條件的直線(xiàn)SKIPIF1<0有(

)A．SKIPIF1<0條 B．SKIPIF1<0條 C．SKIPIF1<0條 D．SKIPIF1<0條【答案】C【解析】直線(xiàn)SKIPIF1<0，即SKIPIF1<0恒過(guò)點(diǎn)SKIPIF1<0，又雙曲線(xiàn)的漸近線(xiàn)方程為SKIPIF1<0，則點(diǎn)SKIPIF1<0在其中一條漸近線(xiàn)SKIPIF1<0上，又直線(xiàn)與雙曲線(xiàn)只有一個(gè)交點(diǎn)，則直線(xiàn)SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0且平行于SKIPIF1<0或過(guò)點(diǎn)SKIPIF1<0且與雙曲線(xiàn)的右支相切，即滿(mǎn)足條件的直線(xiàn)SKIPIF1<0有SKIPIF1<0條.故選：C題型五：雙曲線(xiàn)的弦例8．已知雙曲線(xiàn)SKIPIF1<0：SKIPIF1<0的左右頂點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，點(diǎn)SKIPIF1<0，SKIPIF1<0在雙曲線(xiàn)SKIPIF1<0上．(1)求直線(xiàn)SKIPIF1<0，SKIPIF1<0的斜率之積；(2)若直線(xiàn)MN的斜率為2，且過(guò)點(diǎn)SKIPIF1<0，求SKIPIF1<0的值．【解析】(1)設(shè)SKIPIF1<0，由雙曲線(xiàn)SKIPIF1<0：SKIPIF1<0可得SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，即SKIPIF1<0.(2)直線(xiàn)SKIPIF1<0，設(shè)SKIPIF1<0由SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0.題型六：雙曲線(xiàn)的綜合問(wèn)題例9．已知雙曲線(xiàn)SKIPIF1<0：SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0，漸近線(xiàn)方程為SKIPIF1<0，直線(xiàn)SKIPIF1<0是雙曲線(xiàn)SKIPIF1<0右支的一條切線(xiàn)，且與SKIPIF1<0的漸近線(xiàn)交于A(yíng)，B兩點(diǎn)．(1)求雙曲線(xiàn)SKIPIF1<0的方程；(2)設(shè)點(diǎn)A，B的中點(diǎn)為M，求點(diǎn)M到y(tǒng)軸的距離的最小值．【解析】(1)由題設(shè)可知SKIPIF1<0，解得SKIPIF1<0則SKIPIF1<0：SKIPIF1<0．(2)設(shè)點(diǎn)M的橫坐標(biāo)為SKIPIF1<0當(dāng)直線(xiàn)SKIPIF1<0斜率不存在時(shí)，則直線(xiàn)SKIPIF1<0：SKIPIF1<0易知點(diǎn)SKIPIF1<0到SKIPIF1<0軸的距離為SKIPIF1<0﹔當(dāng)直線(xiàn)SKIPIF1<0斜率存在時(shí)，設(shè)SKIPIF1<0：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，聯(lián)立SKIPIF1<0，整理得SKIPIF1<0，SKIPIF1<0，整理得SKIPIF1<0聯(lián)立SKIPIF1<0，整理得SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0則SKIPIF1<0，即SKIPIF1<0∴此時(shí)點(diǎn)SKIPIF1<0到SKIPIF1<0軸的距離大于2；綜上所述，點(diǎn)SKIPIF1<0到SKIPIF1<0軸的最小距離為2．題型七：直線(xiàn)與拋物線(xiàn)的位置關(guān)系例10．過(guò)拋物線(xiàn)SKIPIF1<0的焦點(diǎn)作一條直線(xiàn)與拋物線(xiàn)交于A(yíng)，B兩點(diǎn)，它們的橫坐標(biāo)之和等于2，則這樣的直線(xiàn)(

)A．有且只有一條B．有且只有兩條C．有且只有三條D．有且只有四條【答案】B【解析】根據(jù)題意，拋物線(xiàn)的焦點(diǎn)坐標(biāo)為SKIPIF1<0，設(shè)SKIPIF1<0，若直線(xiàn)的斜率不存在，則SKIPIF1<0，不符合題意，若直線(xiàn)的斜率存在，設(shè)直線(xiàn)方程為SKIPIF1<0，聯(lián)立SKIPIF1<0，可得SKIPIF1<0，由題意得SKIPIF1<0，解得SKIPIF1<0，所以此時(shí)有兩條直線(xiàn)滿(mǎn)足題意，綜上所述，符合題意得直線(xiàn)有且只有兩條.故選：B.題型八：拋物線(xiàn)的弦例11．過(guò)拋物線(xiàn)SKIPIF1<0：SKIPIF1<0的焦點(diǎn)SKIPIF1<0的直線(xiàn)交拋物SKIPIF1<0線(xiàn)于SKIPIF1<0、SKIPIF1<0兩點(diǎn)，且SKIPIF1<0，則弦SKIPIF1<0的長(zhǎng)為_(kāi)_____.【答案】SKIPIF1<0【解析】由拋物線(xiàn)的焦點(diǎn)弦長(zhǎng)公式可知，SKIPIF1<0.由拋物線(xiàn)方程，可得SKIPIF1<0，又SKIPIF1<0，所以弦SKIPIF1<0的長(zhǎng)為SKIPIF1<0.故答案為：SKIPIF1<0.例12．已知拋物線(xiàn)SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，過(guò)SKIPIF1<0的弦SKIPIF1<0滿(mǎn)足SKIPIF1<0，則SKIPIF1<0的值為_(kāi)_____．【答案】SKIPIF1<0【解析】如圖，由SKIPIF1<0，SKIPIF1<0分別向拋物線(xiàn)的準(zhǔn)線(xiàn)作垂線(xiàn)，垂足為SKIPIF1<0，SKIPIF1<0，設(shè)直線(xiàn)SKIPIF1<0與拋物線(xiàn)的準(zhǔn)線(xiàn)交點(diǎn)為SKIPIF1<0，拋物線(xiàn)的準(zhǔn)線(xiàn)與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，則SKIPIF1<0，設(shè)SKIPIF1<0(SKIPIF1<0)，則SKIPIF1<0，SKIPIF1<0由拋物線(xiàn)的定義，SKIPIF1<0，SKIPIF1<0，易知SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，又易知，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0.故答案為：SKIPIF1<0.題型九：拋物線(xiàn)的綜合問(wèn)題例13．已知拋物線(xiàn)SKIPIF1<0是拋物線(xiàn)SKIPIF1<0上的點(diǎn)，且SKIPIF1<0.(1)求拋物線(xiàn)SKIPIF1<0的方程；(2)已知直線(xiàn)SKIPIF1<0交拋物線(xiàn)SKIPIF1<0于SKIPIF1<0兩點(diǎn)，且SKIPIF1<0的中點(diǎn)為SKIPIF1<0，求直線(xiàn)SKIPIF1<0的方程.【解析】(1)由題意，在拋物線(xiàn)SKIPIF1<0中，SKIPIF1<0，由幾何知識(shí)得，SKIPIF1<0，解得：SKIPIF1<0，故拋物線(xiàn)SKIPIF1<0的方程為：SKIPIF1<0.(2)由題意及(1)得，直線(xiàn)SKIPIF1<0的斜率存在，設(shè)直線(xiàn)SKIPIF1<0的斜率為SKIPIF1<0，則SKIPIF1<0，兩式相減得SKIPIF1<0，整理得SKIPIF1<0，因?yàn)镾KIPIF1<0的中點(diǎn)為SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0，∴直線(xiàn)SKIPIF1<0的方程為：SKIPIF1<0，即SKIPIF1<0，經(jīng)檢驗(yàn)，滿(mǎn)足題意.例14．已知拋物線(xiàn)SKIPIF1<0的焦點(diǎn)為F，點(diǎn)F到拋物線(xiàn)準(zhǔn)線(xiàn)距離為4．(1)求拋物線(xiàn)E的標(biāo)準(zhǔn)方程；(2)已知SKIPIF1<0的三個(gè)頂點(diǎn)都在拋物線(xiàn)E上，頂點(diǎn)SKIPIF1<0，SKIPIF1<0重心恰好是拋物線(xiàn)E的焦點(diǎn)F．求SKIPIF1<0所在的直線(xiàn)方程．【解析】(1)由題意得SKIPIF1<0，∴拋物線(xiàn)方程為：SKIPIF1<0(2)設(shè)SKIPIF1<0，SKIPIF1<0，由重心坐標(biāo)公式得SKIPIF1<0，∴CD中點(diǎn)坐標(biāo)為SKIPIF1<0，SKIPIF1<0兩式相減得SKIPIF1<0，SKIPIF1<0方程：SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0方程：SKIPIF1<0.【過(guò)關(guān)測(cè)試】一、單選題1．已知直線(xiàn)SKIPIF1<0與拋物線(xiàn)SKIPIF1<0交于A(yíng)，B兩點(diǎn)，若D為線(xiàn)段AB的中點(diǎn)，O為坐標(biāo)原點(diǎn)，則直線(xiàn)OD的斜率為(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】設(shè)SKIPIF1<0,則SKIPIF1<0，相減得SKIPIF1<0，由于SKIPIF1<0,所以SKIPIF1<0，所以SKIPIF1<0,將其代入SKIPIF1<0中可得SKIPIF1<0，所以SKIPIF1<0,故SKIPIF1<0,故選：C2．橢圓mx2＋ny2＝1與直線(xiàn)y＝1－x交于M，N兩點(diǎn)，過(guò)原點(diǎn)與線(xiàn)段MN中點(diǎn)的直線(xiàn)的斜率為SKIPIF1<0，則SKIPIF1<0等于()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】設(shè)SKIPIF1<0，M、N中點(diǎn)為D，則SKIPIF1<0，由題意得：SKIPIF1<0因?yàn)镸、N在橢圓上，則SKIPIF1<0，兩式相減整理得SKIPIF1<0，∴SKIPIF1<0.故選：B.3．過(guò)點(diǎn)SKIPIF1<0的直線(xiàn)SKIPIF1<0與雙曲線(xiàn)SKIPIF1<0相交于SKIPIF1<0兩點(diǎn)，若SKIPIF1<0是線(xiàn)段SKIPIF1<0的中點(diǎn)，則直線(xiàn)SKIPIF1<0的方程是(

)A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】設(shè)SKIPIF1<0，則SKIPIF1<0，兩式相減得直線(xiàn)的斜率為SKIPIF1<0，又直線(xiàn)SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0，所以直線(xiàn)SKIPIF1<0的方程為SKIPIF1<0，經(jīng)檢驗(yàn)此時(shí)SKIPIF1<0與雙曲線(xiàn)有兩個(gè)交點(diǎn).故選：A4．在平面直角坐標(biāo)系SKIPIF1<0中，已知點(diǎn)SKIPIF1<0，SKIPIF1<0在橢圓SKIPIF1<0上，且直線(xiàn)SKIPIF1<0，SKIPIF1<0的斜率之積為SKIPIF1<0，則SKIPIF1<0(

)A．1 B．3 C．2 D．SKIPIF1<0【答案】A【解析】因?yàn)辄c(diǎn)SKIPIF1<0，SKIPIF1<0在橢圓SKIPIF1<0上，所以SKIPIF1<0，因?yàn)橹本€(xiàn)SKIPIF1<0，SKIPIF1<0的斜率之積為SKIPIF1<0，所以SKIPIF1<0，可得SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0，則SKIPIF1<0SKIPIF1<0.故選：A.二、填空題5．設(shè)P是雙曲線(xiàn)SKIPIF1<0右支上任一點(diǎn)，過(guò)點(diǎn)P分別作兩條漸近線(xiàn)的垂線(xiàn)，垂足分別為E、F，則SKIPIF1<0的值為_(kāi)_______．【答案】SKIPIF1<0【解析】漸近線(xiàn)方程為SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0．由點(diǎn)到直線(xiàn)的距離公式有SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0.故答案為：SKIPIF1<06．過(guò)拋物線(xiàn)SKIPIF1<0的焦點(diǎn)作一直線(xiàn)交拋物線(xiàn)于SKIPIF1<0、SKIPIF1<0兩點(diǎn)，則SKIPIF1<0的值是________．【答案】SKIPIF1<0【解析】由題意知，拋物線(xiàn)焦點(diǎn)坐標(biāo)為SKIPIF1<0，從而設(shè)直線(xiàn)AB的方程為SKIPIF1<0，聯(lián)立方程SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.所以SKIPIF1<0SKIPIF1<0SKIPIF1<0.故答案為：SKIPIF1<0.7．已知雙曲線(xiàn)C：SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，其中SKIPIF1<0與拋物線(xiàn)SKIPIF1<0的焦點(diǎn)重合，點(diǎn)P在雙曲線(xiàn)C的右支上，若SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的面積為_(kāi)______.【答案】SKIPIF1<0【解析】由雙曲線(xiàn)右焦點(diǎn)SKIPIF1<0與拋物線(xiàn)SKIPIF1<0的焦點(diǎn)重合，可得SKIPIF1<0，所以SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0