22．解：（1）点A的坐标（23，0）；……………… 1分
（2）如图； ………………2分
（3）EF垂直平分OA，
则∠AOD=∠OFE． ∴tan∠AOD =tan∠OFE=
yDE1AC1． 2O1BFx在Rt△AOD中，DA= OD tan∠AOD?3．
∴点A的坐标为?3，···························································································· 3分 6?； ·（4）?1?1····················································································································· 5分 k?? ·3五、解答题（本题共22分，第23题7分，第24题7分，第25题8分） 23．（1）解：∵(k?1)2∴ k∴
··············································································· 1分 ?k(k?1)?3?0， ·
?2.
····································································································· 2分 y?x2?2x?3． ·
（2）由（1），得B (3,0)，C (0,?3)，
∴经过B，C的直线：y·························································································· 3分 ?x?3． ·
设平移后的抛物线G的的顶点M的坐标为：(1,t)． ∵M在线段BC上， ∴t?1?3??2． ···················································································································· 4分
y?(x?1)2?2．
∴平移后的抛物线G为：即
·················································································································· 5分 y?x2?2x?1． ·
（3）作OE⊥BC于点E．
∴B?C??BC?32，OE∴S?OB?C?∴
132BC?． 22119?OC??h?B?C??OE?． 222?5?OC??17．
11
∴
91795?h?．··············································································································· 7分 175
24．（1）解：EG?GC，
EC····················································································· 2分 ?2； ·
GC （2）(1)中的结论仍然成立．
证明：取线段BF的中点M，连接EM，MG，
∵△BEF是等腰直角三角形， ∴EM?MB?FM?1． FB，且∠FME=90°2连接BD，取线段BD的中点N，连接GN，CN， ∵ABCD是正方形， ∴GN?BN?DN?1． BD，且∠CND=90°2∵G是DF的中点， ∴GN?1FB?EM，GN∥FB． 21BD?CN，MG∥BD． 2∴∠1=∠2． 同理MG?∴∠2=∠3． ∴GN?1FB?EM，GN∥FB． 2∴∠1=∠3．
∴∠EMG=∠EMF+∠1=∠CND+∠2=∠GNC．
∴△EMG≌△GNC． ·························································································· 4分 ∴EG=GC． ∴∠EGM=∠GCN．
在△CNG中，∠GNC+∠GCN+∠CGN=180°． ∴∠3+∠GCN+∠CGN=90°． ∴∠2+∠EGM+∠CGN=90°． 即EG⊥GC,
EC···························································································· 5分 ?2． ·
GCA12
（3）当E，F，D三点共线时，连接BD.
FGDE∵BE=1，AB=∴BF2，
?2，BD=2．
?BE1?． BD2在Rt△BED中，sin?EDB∴?EDB?30?． ∴DE?3．
···································································································· 6分 DF?DE?EF?3?1． ·∴?FBD??EFB??EDB∴?ABF?45??30??15?．
??ABD??FBD?45??15??30?．
∴tan?ABF?3． ················································································································· 7分 325．解：（1）
1111, (,?,) ········································································································· 2分 2222（2）答：m?n?k分两种情况：
?S?ABO．
(ⅰ) 当点P在△ABO的内部时，
m?n?k?S?PBO?S?POA?S?PAB?S?PBO?S?POA?S?PAB?S?ABO?S?ABO．
·········································································································································· 3分
(ⅱ) 当点P在△ABO的外部时， yPABP-1OyA2121B1x-1O1xm?n?k?S?PBO?S?POA?S?PAB?S?PBO?S?POA?S?PAB?S?ABO?S?ABO．
综上，m?n?k② P(·························································································· 4分 ?S?ABO. ·
11························································································ 5分 y,?x,1?x?y). ·
22（3）当x>0或x<0时，均有：
13
111y，S?QOA??x，S?QDO?x，S?QOC?y.
2221由S?QBO?S?QOA?S?QAB?S?ABO，得y?x?S?QAB?1.
2111由S?QDO?S?QOC?S?QCD?S?CDO，得x?y?S?QCD?.
222111113S?QAB?S?QCD?(y?x?1)?(x?y?)??x?y?
22222213??x?(x2?2x?4)?
2235?x2?x?
22331?(x?)2?.
41635可以验证当x=0时，y=4，S?QAB?S?QCD?1??，上式仍然成立.
2233∴当S?QAB?S?QCD的值最小时，x??，即点Q的横坐标为?. ······························ 5分
44S?ABO?
QADB-112yyQADB1C21
Ox-1O1Cx 14
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