高考復(fù)習(xí)專題——圓錐曲線背景下的最值與定值問題(參考版)
2024-11-22 22:38本頁面
【正文】 6,43 )2( 。 (2) 當(dāng) a在什么范圍時(shí) , 對(duì)于拋物線上的任意一點(diǎn) M (M與 O不重合 ), ∠ CMF恒為銳角？ ,)(2, 0)(2 :),(),(),( )1( 2221221222222211kpakxxaxxakxpakxkyxByxAaxkyC??????????方程則有將直線方程代入拋物線記的直線為設(shè)過[解析 ] ).(2,0)2(,0,902 )( ))((2212122121221221不存在時(shí)也符合當(dāng)故為正數(shù)由于故時(shí)當(dāng)故kpaaapayyxxA OBapaaxaxxxkaxaxkyy??????????????????1 )0( 02)23(022)21(0))(2(,0 ,),(),2(),0,(),0,2(),()2( 222???????????????????????xapxpaxpxapxpaxyxaxpMCMFC M FyxaMCyxpMFaCpFyxM故恒為銳角但故由于設(shè))29,2()2,0(,2,20 02300292,0495,024)493( 012223pppaCFpapapapapppaaappapa?的范圍是：條件的故知滿足應(yīng)舍去重合和時(shí)但解得：時(shí)，則時(shí)，恒成立，為使???????????????????????????., )3( 。00003?????????????yxdyxxy[解析 ] D .85518121512.161,21.21439。//39。0 )1( 43214321 ?????? kkkkkkkk 且求證：,.,224322212221221212111211122222222122221abkkabybayaxykkaxykaxykybaaxybaax???????????????同理可得：即,2,2,.2:,22,2222431122221111111214321OQBQAQOPBPAPOyxabxxyxabaxyxaxyaxykkkkkk?????????????????則為原點(diǎn)設(shè)同利可得于是.0:)2()1(,.,:43212211??????kkkkyxyxOQOPOQOP得、由于是共線與故由條件知 ?.0:)2()1(,.,:43212211??????kkkkyxyxOQOPOQOP得、由于是共線與故由條件知 ?,1, )2( 22212212222222121?????????????byaxbyaxyyxxOQOP又?.11,.39。),(39。 )2( 242322212222的值求若均為兩曲線的右焦點(diǎn)個(gè)焦點(diǎn)一分別為雙曲線和橢圓的、設(shè)kkkkQFPFFF???。 )1( ).2,1(, 12 22DCBADCABABMABBAyx ??[解析 ] .1012)2(2),(),(,0064)2(2)2(12222122112222???????????????????????????????xyABkkkkxxeyxByxAkkxkkxkyxkkxy：直線滿足則設(shè)時(shí)當(dāng)?shù)茫河?12121212121212121222221212211)(2 ,))((21))((:,1212),(),(yyxxxxyyxxyyyyxxxxyxyxyxByxA????????????????????????兩式相減得則法二：設(shè).012.1:,121222???????????得：代入yxxyABkAB.012.1:,121222???????????得：代入yxxyABkAB[解析 ] 法一為韋達(dá)定理法，法二稱為點(diǎn)差法，當(dāng)涉及到弦的中點(diǎn)時(shí)，常用這兩種途徑處理 . 在利用點(diǎn)差法時(shí)，必須檢驗(yàn)條件△ 0是否成立 . ).4,3(),0,1(:121, : ., )2( 22BAyxxyMDMCMBMAMCDCDMCDCDABMABOMDCBA?????????????得由滿足中點(diǎn)因此只需證點(diǎn)中為故圓心為弦又上垂直平分線即在故為弦因共圓于圓、設(shè)),6,3(63,32),(),(),(0116123,3:00430004433222??????????????????????????MxyxxxyxMCDyxDyxCxxyxxyxyCD則中點(diǎn)設(shè)得：由方程又.102,)6,3(.
教學(xué)課件相關(guān)推薦
鄂ICP備17016276號(hào)-1