教育配套资料K12
即AB=5． ·························· 6
25．（1）4.3； ······························ 1
（2）如图 ····························· 4
（3）3.0或5.2． ··························· 6
26．解：（1）令y=0，得ax2?2ax?3a?0， 解得x1??1，x2=3．
∴A（－1,0），B（3,0）． ···················· 2 （2）∴AB=4．
∵抛物线对称轴为x=1， ∴AM=2． ∵DM=2AM， ∴DM=4．
∴D（1, －4）． ························ 3 ∴a=1．
∴抛物线的表达式为y?x?2x?3． ·············· 4 （3）当∠ADM=45°时，a=
21． ··················· 5 23． 2 当∠ADM=30°时，a=
∴
31
22教育配套资料K12
教育配套资料K12
27．（1）如图 ······························· 1
ADADOFBECOMFEBC
（2）证明：∵BE平分∠CBD，
∴∠CBE=∠DBE． ························ 2 ∵正方形ABCD的对角线AC，BD交于点O， ∴∠BOC=∠BCD=90°． ∵∠CBE+∠CEB=90°， ∠DBE+∠BFO=90°，
∴∠CEB=∠BFO． ························ 3 ∵∠EFC=∠BFO， ∴∠EFC=∠CEB．
∴CF=CE． ··························· 4 （3）证明：取BE的中点M，连接OM． ···················· 5 ∵O为AC的中点，
∴OM∥DE， DE=2OM． ······················ 6 ∴∠OMF=∠CEF．
∵∠OFM=∠EFC=∠CEF， ∴∠OMF=∠OFM． ∴OF=OM． ∴DE=2OF． ··························· 7
28．解：（1）○1P1 ，P2； ························· 2
○2当直线y=x+b与
O相切时，b?22或?22； ········ 3
∴?22?b?22． ···················· 5 （2）当直线y=4与M相切时，m=2或6． ·············· 6 ∴2≤m≤6． ························· 7
教育配套资料K12
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