一道难的初中数学压轴题
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\u5373PM EP =4/ 5 \uff0c\u5219PM=4/ 5 \uff0c
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\u5219MQ=5-3/ 5 =22/ 5 \uff0c
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PQ= (4 5 )2+(22 5 )2 =2 5 \uff1b
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\u2234EP/ FM =EQ/ QF \uff0c\u53732 /FM =6 /4 \uff0c
\u2234FM=4 /3 \uff0c
\u5219MD=4-4/ 3 =8 /3 \uff0cMC=22 /3 \uff0c
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\u2234GQ=t-EG=2/ 5 t+12 /5 \uff0c
PQ2=PG2+GQ2=\uff084/ 5 t-16/ 5 \uff092+\uff082/ 5 t+12 /5 \uff092\uff0c
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\u2234GN=AB=4\uff0c
\u2235M\u662fAD\u7684\u4e2d\u70b9\uff0c
\u2234AM=1\uff0c
\u2234MG/ EM =GN/ MA =4\uff0c
\u2235GM\u22a5EF\uff0c
\u2234\u5728Rt\u25b3EMG\u4e2d\uff0c
\u2234tan\u2220MEG=MG /EM =4\uff1b
\uff082\uff09\u7531\uff081\uff09\u77e5\uff0cMG /EM =4\uff0c\u5373MG=4EM\uff0c
\u2235\u5728Rt\u25b3AEM\u4e2d\uff0cEM= x2+1 \uff0c
\u2234MG=4 x2+1 \uff0c
\u2235S\u25b3EMG=1 2 EM•MG\uff0c
\u2234y=2x2+2 \uff081/ 4 \uff1cx\u22644\uff09\uff1b
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\u2234BE=4-x\uff0cIG=4x\uff0c
\u2234BG=4x+1\uff0cCF=x+4\uff0cCG=4x-1\uff0cCH=2x-1\uff0c
\u2234EF=PG\uff0c\u2220F=\u2220PGC\uff0c
\u2235\u25b3PGC\u223d\u25b3EFQ\uff0c
\u2234\u2220QEF=\u2220CPG\u6216\u2220QEF=\u2220PCG\uff0c
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\u2234QF=CG=4x-1\uff0c
\u2234CQ=CF-QF=5-3x\uff0c
\u53ef\u8bc1BE\u2225CQ\uff0c
\u2234CG BG =CQ BE \uff0c\u5373CG•BE=CQ•BG\uff0c
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\u89e3\u5f97\uff1ax1=3/ 4 2 \uff0cx2= -3/ 4 2 \uff08\u820d\u53bb\uff09\uff0c
\u2234y=17 /4 \uff1b
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\u2234\u70b9H\u5728\u70b9C\u7684\u53f3\u4fa7\uff0c\u5373CH=2x-1\uff0c
\u53c8\u53efPH /CH =tan\u2220MEG=4\uff0c\u5373PH=4CH\uff0c \u22342=4\uff082x-1\uff09\uff0c
\u89e3\u5f97\uff1ax=3/ 4 \uff0c
\u2234y=25/ 8
\u7efc\u4e0a\u6240\u8ff0\uff0c\u53ef\u77e5y\u7684\u503c\u662f17 /4 \u621625/ 8 \uff0e
（2）①EF∥AB． ……………………………………4分
证明：如图，由题意可得A（–4，0），B（0，3），E（-4，-k2/4） ，F（k2/3，3） ．
∴PA=3，PE= ，PB=4，PF= k2/3+4．
∴ PA/PE=12/（12+k）， PB/PF=12/（12+k），
∴ PA/PE=PB/PF． ………………………… 6分
又∵∠APB=∠EPF．
∴△APB ∽△EPF，∴∠PAB=∠PEF．
∴EF∥AB． …………………………… 7分
②S2没有最小值，理由如下：
过E作EM⊥y轴于点M，过F作FN⊥x轴于点N，两线交于点Q．
由上知M（0，-k2/4 ），N（ k2/3，0），Q（ k2/3， -k2/4）． ……………… 8分
而S△EFQ= S△PEF，
∴S2＝S△PEF－S△OEF＝S△EFQ－S△OEF＝S△EOM＋S△FON＋S矩形OMQN
=1/12*(k2+6)2-3 ． ………………………… 10分
当 k2＞-6时，S2的值随k2的增大而增大，而0＜k2＜12． …………… 11分
∴0＜S2＜24，s2没有最小值． …………………………… 12分
说明：1．证明AB∥EF时，还可利用以下三种方法．方法一：分别求出经过A、B两点和经过E、F两点的直线解析式，利用这两个解析式中x的系数相等来证明AB∥EF；方法二：利用 ＝ 来证明AB∥EF；方法三：连接AF、BE，利用S△AEF＝S△BFE得到点A、点B到直线EF的距离相等，再由A、B两点在直线EF同侧可得到AB∥EF．
2．求S2的值时，还可进行如下变形：
S2＝ S△PEF－S△OEF＝S△PEF－（S四边形PEOF－S△PEF）＝2 S△PEF－S四边形PEOF，再利用第（1）题中的结论．
如图解答
如图
给我分啊
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绛旓細璇烽棶鍝噷鐪嬩笉鎳傦紝鍙互杩介棶
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绛旓細2012-09-02 浜,浣犳槸涓嶆槸鏈夈2009骞村叏鍥藉悇鍦颁腑鑰冩暟瀛﹀帇杞撮涓撻泦(鏈榻愬叏... 2009-05-19 鍝釜甯傜殑鏁板涓鍘嬭酱棰樻渶闅 14 2010-02-28 2009鍒濅腑鏁板鍘嗗勾涓冮鍘嬭酱棰樺甫瑙 2013-10-27 鍝釜鍩庡競涓冮鏈闅 51 2012-06-30 鍙蹭笂鏈闅剧殑涓冩暟瀛﹀帇杞撮鏄粈涔? 77 2012-05-26 姹11.10.09骞村叏鍥藉悇鍦颁腑鑰冩暟瀛...
绛旓細鍖楀笀澶х増2010骞涓冨帇杞撮姹囩紪3 濮撳悕___鐝骇___瀛﹀彿___鍒嗘暟___銆愮煡璇嗙旱妯 鍑芥暟(鏈妭涓昏鎸囦竴娆″嚱鏁般佸弽姣斾緥鍑芥暟)鍙婂浘鍍忎笌鍑犱綍闂,鏄互鍑芥暟涓鸿儗鏅帰姹傚嚑浣曟ц川,杩欑被棰樺緢閲嶈鐐规槸鍒╃敤鍑芥暟鐨勬ц川,瑙ｅ喅鍑犱釜涓昏鐐圭殑鍧愭爣闂,浣垮嚑浣曠煡璇嗗拰鍑芥暟鐭ヨ瘑鏈夋満鑰岃嚜鐒剁粨鍚堣捣鏉,杩欐牱,鎵嶈兘绐佺牬闅剧偣｡浣嗗湪...
