(浙江專用)2020版高考數(shù)學(xué)新增分大一輪復(fù)習(xí) 第九章 平面解析幾何 9.5 橢圓(第2課時(shí))直線與橢圓課件.ppt
第2課時(shí)直線與橢圓,第九章9.5橢圓,NEIRONGSUOYIN,內(nèi)容索引,題型分類深度剖析,課時(shí)作業(yè),題型分類深度剖析,1,PARTONE,1.若直線ykx1與橢圓總有公共點(diǎn),則m的取值范圍是A.m>1B.m>0C.0<m0且m5,m1且m5.,將代入,整理得9x28mx2m240.方程根的判別式(8m)249(2m24)8m2144.當(dāng)>0,即時(shí),方程有兩個(gè)不同的實(shí)數(shù)根,可知原方程組有兩組不同的實(shí)數(shù)解.這時(shí)直線l與橢圓C有兩個(gè)不重合的公共點(diǎn).,2.已知直線l:y2xm,橢圓C:試問當(dāng)m取何值時(shí),直線l與橢圓C:(1)有兩個(gè)不重合的公共點(diǎn);,解將直線l的方程與橢圓C的方程聯(lián)立,,(2)有且只有一個(gè)公共點(diǎn);,解當(dāng)0,即m時(shí),方程有兩個(gè)相同的實(shí)數(shù)根,可知原方程組有兩組相同的實(shí)數(shù)解.這時(shí)直線l與橢圓C有兩個(gè)互相重合的公共點(diǎn),即直線l與橢圓C有且只有一個(gè)公共點(diǎn).,(3)沒有公共點(diǎn).,解當(dāng)0.,一般地,在橢圓與向量等知識(shí)的綜合問題中,平面向量只起“背景”或“結(jié)論”的作用,幾乎都不會(huì)在向量的知識(shí)上設(shè)置障礙,所考查的核心內(nèi)容仍然是解析幾何的基本方法和基本思想.,(1)求橢圓C的方程;,解設(shè)A(x1,y1),B(x2,y2),P(x0,y0),,消去y,可得(34k2)x28kmx4m2120,,又點(diǎn)P在橢圓C上,,課時(shí)作業(yè),2,PARTTWO,1.若直線mxny4與O:x2y24沒有交點(diǎn),則過點(diǎn)P(m,n)的直線與橢圓1的交點(diǎn)個(gè)數(shù)是A.至多為1B.2C.1D.0,基礎(chǔ)保分練,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析由題意知橢圓的右焦點(diǎn)F的坐標(biāo)為(1,0),則直線AB的方程為y2x2.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析設(shè)弦的端點(diǎn)A(x1,y1),B(x2,y2),,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,4.已知F1(1,0),F(xiàn)2(1,0)是橢圓C的兩個(gè)焦點(diǎn),過F2且垂直于x軸的直線與橢圓C交于A,B兩點(diǎn),且|AB|3,則C的方程為,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,則c1.因?yàn)檫^F2且垂直于x軸的直線與橢圓交于A,B兩點(diǎn),且|AB|3,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析依題意,當(dāng)直線l經(jīng)過橢圓的右焦點(diǎn)(1,0)時(shí),其方程為y0tan45(x1),即yx1.,A.4B.3C.2D.1,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,PF1PF2,F(xiàn)1PF290.設(shè)|PF1|m,|PF2|n,則mn4,m2n212,2mn4,mn2,,7.直線ykxk1與橢圓1的位置關(guān)系是_.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析由于直線ykxk1k(x1)1過定點(diǎn)(1,1),而(1,1)在橢圓內(nèi),故直線與橢圓必相交.,相交,8.(2018浙江余姚中學(xué)質(zhì)檢)若橢圓C:1的弦被點(diǎn)P(2,1)平分,則這條弦所在的直線l的方程是_,若點(diǎn)M是直線l上一點(diǎn),則M到橢圓C的兩個(gè)焦點(diǎn)的距離之和的最小值為_.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,x2y40,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析當(dāng)直線l的斜率不存在時(shí)不滿足題意,,9.已知橢圓C:1(a>b>0)的左焦點(diǎn)為F,橢圓C與過原點(diǎn)的直線相交于A,B兩點(diǎn),連接AF,BF,若|AB|10,|AF|6,cosABF則橢圓C的離心率e_.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析設(shè)橢圓的右焦點(diǎn)為F1,在ABF中,由余弦定理可解得|BF|8,所以ABF為直角三角形,且AFB90,又因?yàn)樾边匒B的中點(diǎn)為O,所以|OF|c5,連接AF1,因?yàn)锳,B關(guān)于原點(diǎn)對(duì)稱,所以|BF|AF1|8,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,(1)求橢圓E的方程;,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解由題意知,直線AB的斜率存在且不為0,故可設(shè)直線AB的方程為xmy1,設(shè)A(x1,y1),B(x2,y2).,因?yàn)镕1(1,0),,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,(1)求橢圓的標(biāo)準(zhǔn)方程;,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解設(shè)橢圓C的焦距為2c,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,(2)過點(diǎn)P(6,0)的直線l交橢圓于A,B兩點(diǎn),Q是x軸上的點(diǎn),若ABQ是以AB為斜邊的等腰直角三角形,求l的方程.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解設(shè)AB的中點(diǎn)坐標(biāo)為(x0,y0),A(x1,y1),B(x2,y2),,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,所以直線l的方程為x3y60.,技能提升練,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析方法一|OA|OF2|2|OM|,M在橢圓C的短軸上,設(shè)橢圓C的左焦點(diǎn)為F1,連接AF1,,又|AF1|2|AF2|2(2c)2,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,方法二|OA|OF2|2|OM|,M在橢圓C的短軸上,,設(shè)橢圓C的左焦點(diǎn)為F1,連接AF1,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,14.已知橢圓1(a>b>0)短軸的端點(diǎn)為P(0,b),Q(0,b),長(zhǎng)軸的一個(gè)端點(diǎn)為M,AB為經(jīng)過橢圓中心且不在坐標(biāo)軸上的一條弦,若PA,PB的斜率之積等于則點(diǎn)P到直線QM的距離為_.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析設(shè)A(x0,y0),則B點(diǎn)坐標(biāo)為(x0,y0),,則直線QM的方程為bxayab0,,拓展沖刺練,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析設(shè)AB的中點(diǎn)為G,則由橢圓的對(duì)稱性知,O為平行四邊形ABCD的對(duì)角線的交點(diǎn),則GOAD.設(shè)A(x1,y1),B(x2,y2),,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解設(shè)M(x0,y0),P(x1,y1),Q(x2,y2),由題意知PQ的斜率存在,且不為0,所以x0y00,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,