【正文】
, 13132sin ?? 32),0(,21c o s0s i n21s i nc o s),s i n (s i n21c o ss i ns i ns i n21c o ss i n,21c o s)1.(20?? ??????????????????AAACCACACCABCCAbcCa?? )3321,2(]1,23()3s i n (),32,3(3),3,0()3s i n (3321)]3s i n ([ s i n3321s i n332s i n3321s i n332,s i n332,s i ns i n)2(?????????????????????????lBBBBBBCBlCcBbBbAa?????????周長同理解法二,可用均值不等式,略 141,3,2,324222 3)1.(2122222?????????????yxcabcacaac橢圓方程為且 2212212222222211221145,4k65080160564143),1(),1(),(),(,303)21(),21(),20(),20(0011)2(kxxkxxkkkxxkyxkxyyxNByxNAyxByxAkxylklNBNANBNABAlNBNAN????????????????????????????????????????得由)得(和聯(lián)立設方程為,設斜率存在時,設斜率為當,不符合條件,斜率不存在時,當)且根據(jù)題意知,()知由( 3533,530445631)(44436)3)(3(22212121222121??????????????????????????????xyxylkkkkkyyxxxxNBNAkkkxkxyy或方程為均滿足或 22.(1) 1,1 )1(11)( 239。 xfxf 當 時10 ??a 由 axaxxf ????????? 1,110)( 2139。 ???? xxg? 命題得證)單調遞增,在( ????? 0)0()(1,0)( gxgxg