《(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 9.6 雙曲線課件.ppt》由會(huì)員分享,可在線閱讀,更多相關(guān)《(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 9.6 雙曲線課件.ppt(100頁(yè)珍藏版)》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。
9.6雙曲線,,第九章平面解析幾何,,NEIRONGSUOYIN,內(nèi)容索引,基礎(chǔ)知識(shí)自主學(xué)習(xí),題型分類深度剖析,課時(shí)作業(yè),1,基礎(chǔ)知識(shí)自主學(xué)習(xí),PARTONE,,知識(shí)梳理,1.雙曲線定義平面內(nèi)與兩個(gè)定點(diǎn)F1,F(xiàn)2的等于常數(shù)(小于|F1F2|)的點(diǎn)的軌跡叫做雙曲線.這兩個(gè)定點(diǎn)叫做,兩焦點(diǎn)間的距離叫做______________.集合P={M|||MF1|-|MF2||=2a},|F1F2|=2c,其中a,c為常數(shù)且a>0,c>0.(1)當(dāng)時(shí),P點(diǎn)的軌跡是雙曲線;(2)當(dāng)時(shí),P點(diǎn)的軌跡是兩條射線;(3)當(dāng)時(shí),P點(diǎn)不存在.,ZHISHISHULI,距離的差的絕對(duì)值,,,,雙曲線的焦點(diǎn),雙曲線,的焦距,2a|F1F2|,2.雙曲線的標(biāo)準(zhǔn)方程和幾何性質(zhì),x≥a或x≤-a,y∈R,x∈R,y≤-a或y≥a,坐標(biāo)軸,原點(diǎn),2.雙曲線的標(biāo)準(zhǔn)方程和幾何性質(zhì),x≥a或x≤-a,y∈R,x∈R,y≤-a或y≥a,坐標(biāo)軸,原點(diǎn),(1,+∞),2a,2b,a2+b2,【概念方法微思考】,1.平面內(nèi)與兩定點(diǎn)F1,F(xiàn)2的距離之差的絕對(duì)值等于常數(shù)2a的動(dòng)點(diǎn)的軌跡一定為雙曲線嗎?為什么?,提示不一定.當(dāng)2a=|F1F2|時(shí),動(dòng)點(diǎn)的軌跡是兩條射線;當(dāng)2a>|F1F2|時(shí),動(dòng)點(diǎn)的軌跡不存在;當(dāng)2a=0時(shí),動(dòng)點(diǎn)的軌跡是線段F1F2的中垂線.,2.方程Ax2+By2=1表示雙曲線的充要條件是什么?,提示若A>0,B0,表示焦點(diǎn)在y軸上的雙曲線.所以Ax2+By2=1表示雙曲線的充要條件是AB0,b>0,二者沒有大小要求,若a>b>0,a=b>0,0
0,b>0)的左、右焦點(diǎn)分別為F1,F(xiàn)2,M是雙曲線C的一條漸近線上的點(diǎn),且OM⊥MF2,O為坐標(biāo)原點(diǎn),若=16,則雙曲線的實(shí)軸長(zhǎng)是A.32B.16C.84D.4,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,所以雙曲線C的實(shí)軸長(zhǎng)為16.故選B.,,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,由雙曲線C1的一條漸近線與圓C2有兩個(gè)不同的交點(diǎn),,又知b2=c2-a2,所以c2>4(c2-a2),,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,9.雙曲線的焦點(diǎn)在x軸上,實(shí)軸長(zhǎng)為4,離心率為則該雙曲線的標(biāo)準(zhǔn)方程為__________,漸近線方程為__________.,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,4,10.已知F1,F(xiàn)2分別是雙曲線x2-=1(b>0)的左、右焦點(diǎn),A是雙曲線上在第一象限內(nèi)的點(diǎn),若|AF2|=2且∠F1AF2=45,延長(zhǎng)AF2交雙曲線的右支于點(diǎn)B,則△F1AB的面積等于____.,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析由題意知a=1,由雙曲線定義知|AF1|-|AF2|=2a=2,|BF1|-|BF2|=2a=2,∴|AF1|=2+|AF2|=4,|BF1|=2+|BF2|.由題意知|AB|=|AF2|+|BF2|=2+|BF2|,∴|BA|=|BF1|,∵△BAF1為等腰三角形,∠F1AF2=45,∴∠ABF1=90,∴△BAF1為等腰直角三角形.,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,(0,2),,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)左焦點(diǎn)為F1,由于雙曲線和圓都關(guān)于x軸對(duì)稱,又△APQ的一個(gè)內(nèi)角為60,∴∠PAF=30,∠PFA=120,|AF|=|PF|=c+a,∴|PF1|=3a+c,在△PFF1中,由余弦定理得|PF1|2=|PF|2+|F1F|2-2|PF||F1F|cos∠F1FP,即3c2-ac-4a2=0,即3e2-e-4=0,,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,解析設(shè)內(nèi)切圓的圓心M(x,y),圓M分別切AF1,BF1,AB于S,T,Q,如圖,連接MS,MT,MF1,MQ,則|F1T|=|F1S|,故四邊形SF1TM是正方形,邊長(zhǎng)為圓M的半徑.由|AS|=|AQ|,|BT|=|BQ|,得|AF1|-|AQ|=|SF1|=|TF1|=|BF1|-|BQ|,又|AF1|-|AF2|=|BF1|-|BF2|,∴Q與F2重合,∴|SF1|=|AF1|-|AF2|=2a,∴|MF2|=2a,即(x-c)2+y2=4a2,①,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,14.如圖,已知F1,F(xiàn)2分別是雙曲線x2-=1(b>0)的左、右焦點(diǎn),過點(diǎn)F1的直線與圓x2+y2=1相切于點(diǎn)T,與雙曲線的左、右兩支分別交于A,B,若|F2B|=|AB|,求b的值.,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解方法一因?yàn)閨F2B|=|AB|,所以結(jié)合雙曲線的定義,得|AF1|=|BF1|-|AB|=|BF1|-|BF2|=2,連接OT,在Rt△OTF1中,|OT|=1,|OF1|=c,|TF1|=b,,化簡(jiǎn)得b6-4b5+5b4-4b3-4=0,得(b2-2b-2)(b4-2b3+3b2-2b+2)=0,而b4-2b3+3b2-2b+2=b2(b-1)2+b2+1+(b-1)2>0,故b2-2b-2=0,,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,方法二因?yàn)閨F2B|=|AB|,所以結(jié)合雙曲線的定義,得|AF1|=|BF1|-|AB|=|BF1|-|BF2|=2,連接AF2,則|AF2|=2+|AF1|=4.連接OT,在Rt△OTF1中,|OT|=1,|OF1|=c,|TF1|=b,,所以c2-3=2b,又在雙曲線中,c2=1+b2,,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,解析方法一當(dāng)A在第一象限時(shí),如圖1,設(shè)△AF1F2的內(nèi)切圓⊙O1分別切AF1,F(xiàn)1F2,F(xiàn)2A于點(diǎn)Q,P,N,則|AQ|=|AN|,|F1Q|=|F1P|,|F2P|=|F2N|,又|AF1|-|AF2|=2a,即(|AQ|+|F1Q|)-(|AN|+|F2N|)=2a,∴|F1Q|-|F2N|=2a,∴|F1F2|-|F2P|-|F2N|=2a,即2c-2|F2P|=2a,∴|F2P|=c-a,∴P為雙曲線的右頂點(diǎn),同理,△BF1F2的內(nèi)切圓⊙O2也切F1F2于雙曲線的右頂點(diǎn),,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,連接O1P,O2P,則O1,P,O2三點(diǎn)共線,且O1O2⊥F1F2.連接O1F2,O2F2,又O1F2平分∠F1F2A,O2F2平分∠F1F2B,∴∠O1F2O2=90,∴Rt△O1F2P∽R(shí)t△F2O2P∽R(shí)t△O1O2F2,∴|O1F2|2=|O1P||O1O2|,|O2F2|2=|O2P||O1O2|,,∴∠O2O1F2=30,,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,則∠O1F2P=60,∴∠AF2P=120,,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,方法二當(dāng)A在第一象限時(shí),如圖2,設(shè)△AF1F2的內(nèi)切圓⊙O1分別切AF1,F(xiàn)1F2,F(xiàn)2A于點(diǎn)Q,P,N,則|AQ|=|AN|,|F1Q|=|F1P|,|F2P|=|F2N|,又|AF1|-|AF2|=2a,即(|AQ|+|F1Q|)-(|AN|+|F2N|)=2a,∴|F1Q|-|F2N|=2a,∴|F1F2|-|F2P|-|F2N|=2a,即2c-2|F2P|=2a,∴|F2P|=c-a,∴P為雙曲線的右頂點(diǎn),同理,△BF1F2的內(nèi)切圓⊙O2也切F1F2于雙曲線的右頂點(diǎn),連接O1P,O2P,則O1,P,O2三點(diǎn)共線,且O1O2⊥F1F2.,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,設(shè)⊙O2切BF2于點(diǎn)H,連接O1N,O2H,則在直角梯形O2HNO1中,|O2H|=r2,|O1N|=r1=3r2,|O1O2|=r1+r2=4r2,作O2T⊥O1N于點(diǎn)T,則|O1T|=r1-r2=2r2,故在Rt△O1O2T中,∠O2O1T=60,,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,,在△PF1F2中,由余弦定理得|PF1|2=|PF2|2+|F1F2|2-2|PF2||F1F2|cos∠PF2F1=c2+(2c)2-2c2ccos60=3c2,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,,因?yàn)橹本€PF2的傾斜角為120,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,整理得c4-8c2a2+4a4=0,即e4-8e2+4=0,,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,
鏈接地址:http://appdesigncorp.com/p-3198724.html