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《全国中考数学压轴题精选3含答案.pdf》由会员上传分享,免费在线阅读,更多相关内容在工程资料-天天文库。
1、2008年全国中考数学压轴题精选精析(三)1921(08江西南昌24题)如图,抛物线yax2ax1经过点P,,且与抛物线yax2ax1相交于1282A,B两点.(1)求a值;(2)设yax2ax1与x轴分别交于M,N两点(点M在点N的左边),yax2ax1与x轴分别12交于E,F两点(点E在点F的左边),观察M,N,E,F四点的坐标,写出一条正确的结论,并通过计算说明;(3)设A,B两点的横坐标分别记为x,x,若在x轴上有一动点Q(x,0),且x≤x≤x,过Q作一条垂ABAB直于x轴的直线,与两条抛物线分别交于C,D
2、两点,试问当x为何值时,线段CD有最大值?其最大值为多少?yPAxOB19(08江西南昌24题解析)解:(1)Q点P,在抛物线yax2ax1上,281119aa1,···························2分4281解得a.································3分211111(2)由(1)知a,抛物线yx2x1,yx2x1.···5分212222211当x2x10时,解得x2,x1.y2212PQ点M在点N的左边,x2,x1.·····6分AMNx11MEONF当x2x10时
3、,解得x1,x2.B2234Q点E在点F的左边,x1,x2.·················7分EFQxx0,xx0,MFNE点M与点F对称,点N与点E对称.···················8分1(3)Qa0.y2P抛物线y开口向下,抛物线y开口向上.······9分AC12xOQBD根据题意,得CDyy121111x2x1x2x1x22.··············11分2222Qx≤x≤x,当x0时,CD有最大值2.··············12分AB
4、说明:第(2)问中,结论写成“M,N,E,F四点横坐标的代数和为0”或“MNEF”均得1分.22(08江西南昌25题)如图1,正方形ABCD和正三角形EFG的边长都为1,点E,F分别在线段AB,AD上滑动,设点G到CD的距离为x,到BC的距离为y,记HEF为(当点E,F分别与B,A重合时,记0o).(1)当0o时(如图2所示),求x,y的值(结果保留根号);(2)当为何值时,点G落在对角形AC上?请说出你的理由,并求出此时x,y的值(结果保留根号);(3)请你补充完成下表(精确到0.01):0o15
5、o30o45o60o75o90ox0.0300.29y0.290.130.03(4)若将“点E,F分别在线段AB,AD上滑动”改为“点E,F分别在正方形ABCD边上滑动”.当滑动一周时,请使用(3)的结果,在图4中描出部分点后,勾画出点G运动所形成的大致图形.6262(参考数据:3≈1.732,sin15o≈0.259,sin75o≈0.966.)44HADHHHFADA(F)DADGGEBCB(E)CBCBC图1图2图3图4(08江西南昌25题解析)解:(1)过G作MNAB于M交CD于N,GKBC于K.
6、QABG60o,BG1,31MG,BM.··························2分2231x1,y.···························3分22(2)当45o时,点G在对角线AC上,其理由是:·············4分H过G作IQ∥BC交AB,CD于I,Q,DA(F)过G作JP∥AB交AD,BC于J,P.GMNQAC平分BCD,GPGQ,GIGJ.B(E)CQGEGF,Rt△GEI≌Rt△GFJ,GEIGFJ.KQGEFGFE6
7、0o,AEFAFE.QEAF90o,AEFAFE45o.即45o时,点G落在对角线AC上.····················6分(以下给出两种求x,y的解法)HFJDA方法一:QAEG45o60o105o,GEI75o.62EGQ在Rt△GEI中,GIGEgsin75o,I4PCB62GQIQGI1.······················7分462xy1.··························8分4方法二:当点G在对角线AC
8、上时,有132x2,···························7分2262解得x1462xy1.··························8分4(3)0o15o30o45o60o75o90ox0.130.0300.030.130.290.50y0.500.290.130.0300.030.13···
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