2009年广东省梅州市中考数学试题及答案(word版)(4)

2

整理得x 5x 9 0． ······································································································· 8 分 ∵b 4ac 25 36 11 0， ∴该方程无实数根．

∴点E不存在，BE不能与⊙O相切． ·········································· 10分 法二： 若BE能与⊙O相切，因AE是⊙O的直径，则AE⊥BE， AEB 90°，

222

设DE x，则EC 5 x，由勾股定理得：AE EB AB，

2

即(9 x2) [(5 x)2 9] 25， 整理得x 5x 9 0， ······································· 8分 ∵b 4ac 25 36 11 0， ∴该方程无实数根．

∴点E不存在，BE不能与⊙O相切． ·········································· 10分 （法三：本题可以通过判断以AB为直径的圆与DC是否有交点来求解，参照前一解法给分） 23．本题满分 11 分．

（1）y 1 x ························································································································ 2分 （2）∵OP t，∴Q点的横坐标为①当0

2

2

1

t， 2

11t 1，即0 t 2时，QM 1 t， 22

∴S△OPQ

1 1

····································································································· 3分 t 1 t ． ·

2 2

11

t t 1， 22

②当t≥2时，QM

∴S△OPQ

1 1

t t 1 ． 2 2

1 1

0 t 2， 2t 1 2t ，

∴S ······························································································ 4分

1t 1t 1 ，t≥2.

22

当0

11 1 11t 1，即0 t 2时，S t 1 t (t 1)2 ， 22 2 44

1

． ······························································································ 6分 4

∴当t 1时，S有最大值

（3）由OA OB 1，所以△OAB是等腰直角三角形，若在L1上存在点C，使得△CPQ

是以Q为直角顶点的等腰直角三角形，则PQ QC，所以OQ QC，又L1∥x轴，则C，

O两点关于直线L对称，所以AC OA 1，得C(11)，． ················································ 7 分

下证 PQC 90°．连CB，则四边形OACB是正方形．

法一：（i）当点P在线段OB上，Q在线段AB上 （Q与B、C不重合）时，如图–1．

由对称性，得 BCQ QOP， QPO QOP， ∴ QPB QCB QPB QPO 180°，

∴ PQC 360° ( QPB QCB PBC) 90°． ················································· 8分 （ii）当点P在线段OB的延长线上，Q在线段AB上时，如图–2，如图–3