【文档说明】内蒙古自治区赤峰市2023-2024学年高二下学期7月期末考试 数学 PDF版含答案.pdf，共(10)页，552.359 KB，由管理员店铺上传

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{#{QQABDYoEogAAAJJAAQhCEQWoCgCQkBECCYgOxFAEoAABABFABAA=}#}{#{QQABDYoEogAAAJJAAQhCEQWoCgCQkBECCYgOxFAEoAABABFABAA=

}#}{#{QQABDYoEogAAAJJAAQhCEQWoCgCQkBECCYgOxFAEoAABABFABAA=}#}{#{QQABDYoEogAAAJJAAQhCEQWoCgCQkBECCYgOxFAEoAABABFABAA=}#}2024年

赤峰市高二年级学年联考数学答案2024．07一、单选题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的．12345678BACACDBD二、多项选择题：本题共4小题，每小题5分，共20分．在每小题给出的选项中，有多项符合

题目要求．全部选对的得5分，部分选对的得2分，有选错的得0分．9101112BDABCADBCD三、填空题：本题共4小题，每小题5分，满分20分．13、114、415、3x−（答案不唯一）16、432,3四、解答题：本题共6小题，共70分．解答应写出必要的文字说明、证明过程或

演算步骤．17.解:(1)设圆C的半径为r，则22131022r=−+−,∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙1分1∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙2分则圆C的方程为：()2211xy−+=.∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙3分(2)因为圆C的半径为1，所以当直线l与圆相交所得的弦长为3时，圆心C到直线l的距离为12∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙4分当直线l的斜率不存在时，直线l：12x=，此时圆心C到直线l的距离为12,满足题意∙∙∙5分当直线l的斜率存在时，{#{QQABDYoEogAAAJJAAQh

CEQWoCgCQkBECCYgOxFAEoAABABFABAA=}#}设直线l：31()22ykx−=−，即2230kxyk−+−=①∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙6分

则22|203|12(2)(2)kkk++−=+−，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙7分解得33k=−，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙8分代入①得：320xy+−=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙9分综上，直线l

的方程为12x=或320xy+−=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙10分18.(1)121nnaa1122nnaa+−=−∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙1分则1122211nnnnaaaa∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙2分12a111a∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙3分{}1na∴−是以1为首项2为公比的等比数列.∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙4分(2)∴112nna−−=∴121nna−=+∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙5分nnbna=⋅即12nnbnn−=⋅+，∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙6分设12nncn−=⋅的前n项和nS令01112222nnSn−=⋅+⋅++⋅①∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙7分2

nS=1212222nn⋅+⋅++⋅②∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙8分②得012122222nnnSn∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙9分∴21nnS（n-1）=+∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙10分()11232nnn+++++=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙11分∴()212nnnT=++n+1（n-

1）∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙12分{#{QQABDYoEogAAAJJAAQhCEQWoCgCQkBECCYgOxFAEoAABABFAB

AA=}#}19.（1）以D为坐标原点以DA所在直线为x轴，DC所在直线为y轴，1DD所在直线为z轴如图建立空间直角坐标系∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙1分设AEx=1(0)2x<≤，则2BFx=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙2分则1(1,0,2)A、(12,2,0)Fx−、1(0,2,2)C、(1,,0)Ex1(2,2,2)AFx=−−、1(1,2,2)CEx=−−

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙3分1122440AFCExx⋅=−+−+=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙4分所以，11AFCE⊥∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙5分（2）1123BBEFBEFVS−∆=××，当BEFS∆最大时，三棱锥1BBEF−的体积最大∙∙∙∙∙∙∙∙∙6分()2112222

2BEFSBEBFxxxx∆=⋅=−⋅=−+1(0)2x<≤当12x=时BEFS∆最大，此时F与C重合，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙7分1(1,2,2)B、

(0,2,0)F、1(1,,0)2E13(0,,2)2BE=−−、3(1,0)2FE=−∙∙∙∙∙8分设面1BEF的法向量()1,,nxyz=11100nBEnFE⋅=⋅=133,2,2

n=−∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙9分面BEF的法向量()20,0,1n=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙10分1212361cos61nnnnθ⋅==⋅∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙11分所以平面1BEF与平面BEF的夹角余弦值为36161∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙12分{#{QQABDYoEogAA

AJJAAQhCEQWoCgCQkBECCYgOxFAEoAABABFABAA=}#}20.（1）用A表示事件该省大学生每天玩手机超过1小时，用B表示事件该省大学生近视，则1(A)5P=，1|2PBA

，3|8PBA，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙1分||PBPBAPAPBAPA∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙2分11312125855∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙3分(2)由

（1）可知，该省近视的大学生中玩手机超过1小时的概率为|PAB∙∙∙∙∙∙∙∙∙∙∙∙∙4分11|125|245PABPBAPAPABPBPB∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙5分所以

该省近视的大学生中玩手机超过1小时的概率为14；∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙6分（3）设抽取人的二人中近视的人数为X，X的所有可能取值为0,1,2∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙7分()()202010.40.36PXC==⋅−=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙8分()()1210.410.40.48PXC==⋅⋅−=∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙9分()22220.40.16PXC==⋅=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙10分X的分布列为X012P0.360.480.16∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙11分()00.3610.4820.160.8EX=×+×+×=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙12分{#{QQABDYoEogAAAJJAAQhCEQWoCgCQkBECCYgOxFAEoAABABFABAA=}#}21．（1）由31()2xfxxex=⋅−，得()00f=，

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙1分对fx求导得()23()12xfxxex′=+−，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙2分∴()01f′=，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙3分∴()fx在()()0,0f处的切线方程为yx=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙4分（2）当1x≥时，()2fxkxx≥+恒成立即1x≥3212xxexkxx⋅−≥+恒成立∴2112xexkx−−≤在[)1+∞，恒成立∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙5分令()2112xexgxx−−=则()()221112xxexgxx−−+′=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙6分令()()21112xmxxex=−−+则()()1xmxxe′=−，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙7分1x≥∴()0mx′>恒成立∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙8分∴当1x≥时，()()21112xmxxex=−−+单调递增∴()()1102mxm≥=>∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙9分∴当1x≥时，()()2211120xxexgxx−−+′=>∙∴当1x≥时，()211

2xexgxx−−=单调递增∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙10分∴()()312gxge≥=−∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙11分∴当1x≥时,32ke≤−∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙12分{#{QQABDYoEogAAAJJAAQhCEQWoCgCQkBECCYgOxFAEoAABABFABAA=}#}22.（1）依题意，由椭圆定义可知P在以()1,0−、()1,0为焦点，25为长轴长的椭圆上∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙1分5a=，1c=，222bac=−=，∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙2分∴曲线C的标准方程为22154xy+=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙4分设()00,Pxy当过P的切线斜率不存在时，不能和直线5x=相交，所以过P的切线l斜率存在，设切线的斜率为k，y轴上的截距为m，则切线:lykxm=+∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

∙∙∙∙∙∙∙∙∙∙∙∙∙5分联立22154xyykxm+==+消去y得()22254105200kxkmxm+++−=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙6分l与曲线C相切，()()()222104545200km

km∴∆=−+−=2254km∴=−∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙7分()02105254kmkxmk=−=−+00544kykxmkmm=+=−+=54,kPmm∴−5xykxm=

=+联立求得Q()5,5km+∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙8分由椭圆的对称性易知

，若过定点则该定点一定在x轴上，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙9分设(),0Mt是以PQ为直径的圆上的一点∴54,kMPtmm=−−()5,5MQtkm=

−+0MPMQ∴⋅=，即()()54550kttkmmm−−−++=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙10分()255540

ktttm−+−+=2550540ttt−=∴−+=1t∴=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙11分∴以PQ为直径的圆过定点(

)1,0M∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙12分{#{QQABDYoEogAAAJJAAQhCEQWoCgCQkBECCYgOxFAEoAABA

BFABAA=}#}