【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,由小赞的店铺上传
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{#{QQABDYoEogAAAJJAAQhCEQWoCgCQkBECCYgOxFAEoAABABFABAA=}#}{#{QQABDYoEogAAAJJAAQhCEQWoCgCQkBECCYgOxFAEoAA
BABFABAA=}#}{#{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,{#{QQABDYoEogAAAJJAAQhCEQWoCgCQkBECCYgOxFAEoAABABFABAA=}#}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:320
xy+−=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙9分综上,直线l的方程
为12x=或320xy+−=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙10分18.(1)121nnaa1122nnaa+−=−∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙
∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙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2nS=1212222nn⋅+⋅++⋅②∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙8分②得012122222nnnSn∙∙∙∙∙∙∙∙∙∙∙∙
∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙9分∴21nnS(n-1)=+∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙
∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙10分()11232nnn+++++=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙
∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙11分∴()212nnnT=++n+1(n-1)∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙
∙∙∙∙12分{#{QQABDYoEogAAAJJAAQhCEQWoCgCQkBECCYgOxFAEoAABABFABAA=}#}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()21122222BEFSBEBFxxxx∆=⋅=−⋅=−+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,
2n=−∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙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{#{QQABDYoEogAAAJJAAQhCEQWoCgCQkBECCYgOxFAEoAABABFABAA=}
#}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,()2112
xexgxx−−=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=,2
22bac=−=,∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙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,()(
)()222104545200kmkm∴∆=−+−=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()255540ktttm−+−+=2
550540ttt−=∴−+=1t∴=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙11分∴以PQ为直径的圆
过定点()1,0M∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙12分{#{QQABDYoEogAAAJJAAQhCEQWoCgCQkBECCYgOxFAEoAABABF
ABAA=}#}