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2025—2026学年度第一学期福九联盟（高中）期中联考
高二数学试卷
完卷时间：120 分钟 满分：150 分
第I卷
一、单项选择题：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．
1. 直线 的一个方向向量为（ ）．
A．(2,−1) B．(2, 1) C．(1,−2) D．(1, 2)
2. 两平行直线 与 的距离为（ ）．
A． B． C． D．
3. 已知椭圆 的一个焦点为 ，则 （ ）．
A． B． C． D．
4. 已知 ， ， ，若 ， ， 三向量共面，则实数 （
）．
A．3 B． C．4 D．
5. 已知圆 ，点 是圆 上一动点，点 ， 为线段 的中点，
则动点 的轨迹方程为（ ）．
A． B．
C． D．
6. 已知椭圆 以及椭圆内一点 ，以点 为中点的弦所在直线的斜率为（
）．
A． B． C． D．2
7. 在平面直角坐标系中，与点
A(1,−2)距离为
1，且与点
B(−2, 2)距离为
2 的直线共有（
）．
高二数学试卷（第1页 共16页）
学学科科网网（（北北京京））股股份份有有限限公公司司A．1条 B．2条 C．3条 D．4条
8. 已知圆 与椭圆 ，若在椭圆 上存在一点 ，过
作圆 的两条切线，切点分别为 ，且 ，则椭圆 离心率的取值范围为
（ ）．
A． B． C． D．
二、多项选择题：本题共3小题，每小题6分，共18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6分，部分选对的得部分分，有选错的得0分．
9. 关于空间向量，以下说法正确的是（ ）．
A．任意向量 ，满足
B．若三个非零空间向量 满足 ，则有
C．若直线l的一个方向向量为 ，平面 的一个法向量为 ，则l⊥α
D．若空间向量 ，则 在 上的投影向量为
10. 已知圆 ，直线 ，则（ ）．
A．直线 恒过定点
B．直线 与圆 有两个交点
C．当 时，圆 上恰有四个点到直线 的距离等于1
D．直线 与圆 相交得到的最短弦长为
11. 已知正方体 的棱长为2， ， 分别是线段 ， 上的动点，且满
足 ， 是线段 的中点，则（ ）．
A．若 是 的中点，则 平面
B．若 是 的中点，则 平面
C． 的最大值是
D． 的最小值为
第Ⅱ卷
高二数学试卷（第2页 共16页）
学学科科网网（（北北京京））股股份份有有限限公公司司三、填空题：本题共3小题，每小题5分，共15分．
12. 在空间直角坐标系Oxyz中，点 关于 平面的对称点坐标是 ．
13. 已知椭圆 的左焦点为 ， 是 上关于原点对称的两点，且
，则 的周长为 ．
14. 人教A版选择性必修一习题 1.4拓广探索第17题中提到“在空间直角坐标系 中，
已知向量 ，点 ，若平面 经过点 ，且以 为法向量，点
是平面内的任意一点，则平面 的方程为 ”．
现已知平面 的方程为 ，直线l是平面 与平面 的
交线，则平面 的一个法向量可以为 （只需写出一个满足条件的向量即
可），直线l与平面 所成角的正弦值为 ．
四、解答题：本题共5小题，共77分．解答应写出文字说明，证明过程或演算步骤．
15. （13分）
已知直线 的方程为 ，若直线 在y轴上的截距为 ，且 ．
（1）求直线 和 的交点坐标；
（2）已知直线 经过 与 的交点，且与两坐标轴的正半轴围成的三角形面积为3，求直线
的方程．
16. （15分）
如图，在正四面体 中， ， 为棱 的中点，
为棱 （靠近 点）的三等分点，设 ．
高二数学试卷（第3页 共16页）
学学科科网网（（北北京京））股股份份有有限限公公司司（1）用 表示 ；
（2）求 ；
（3）求 的长．
17. （15分）
已知圆 经过
A(0,−1)和 B(2,1)，且圆心在直线
上．
（1）求圆 的方程；
（2）若直线l过点(2, 2)，与圆
交于M，N两点， ，求直线l的方程．
18. （15分）
如图（1），在直角梯形 中， ，过 的中点 作 交
于点 ， ，现将四边形 沿着 翻折至 位置，使得
，如图（2）所示．
高二数学试卷（第4页 共16页）
学学科科网网（（北北京京））股股份份有有限限公公司司（1） （2）
（1）证明： 平面 ；
（2）在线段 上是否存在一点 ，使得平面 与平面 的夹角的余弦值为 ，若
存在，确定点 的位置，若不存在，请说明理由．
19. （17分）
A ，A （−2，0）,(2,0)
已知 1 2两点的坐标分别为 ，直线 相交于点 ，它们的斜率之积是
1
−
4．
（1）求点 的轨迹 的方程；
（2）过点 的直线 与 交于 两点．
(ⅰ)求 的取值范围；
(ⅱ)若直线 分别与直线 相交于 两点，求 的值．
高二数学试卷（第5页 共16页）
学学科科网网（（北北京京））股股份份有有限限公公司司2025—2026学年度第一学期福九联盟（高中）期中联考
高中二年数学科试卷评分细则
第I卷
选择题: 单选8题，每小题 5分；多项选择 3题，全部选对的得 6分，部分选对的得部分
分，有选错的得0分．
题号 1 2 3 4 5 6 7 8 9 10 11
答案 C B D D C A D B CD ABD ACD
第Ⅱ卷
三、填空题：本大题共3小题，每小题5分．
2. 13. 14 14. （答案不唯一， 均可），
1
四、解答题：解答应写出文字说明、证明过程或演算步骤．（共5题，13分+15分+15分
+17分+17分，共77分）
15.【解析】（方法一）（1）因为直线 的斜率 且 ，所以直线 的斜率 ，
·············································································································································································2分
因为 在 轴上的截距为 ，
所以直线 方程： ，即 ，························································································3分
高二数学试卷（第6页 共16页）
学学科科网网（（北北京京））股股份份有有限限公公司司联立方程得： ，解得 ，··················································································5分
故交点为 ．··········································································································································6分
（2）依据题意可知：直线 的斜率 ，设直线 ： ，·······························7分
与两坐标轴的交点为 ，···················································································9分
则直线 与两坐标轴的正半轴围成的三角形面积 ，··················10分
解得 ，··············································································································································12分
故 方程为： ，即 . ······································································13分
（方法二）（1）同方法一．···················································································································6分
（2）依据题意可知：直线 的截距存在且不为 ，设直线 ： …7分
在直线 上， ······································································································8分
则直线 与两坐标轴的正半轴围成的三角形面积 ， ··············································9分
联立方程得： ， ···············································································································11分
又 解得 ， ···············································································································12分
故 方程为： ，即 . ·····················································································13分
高二数学试卷（第7页 共16页）
学学科科网网（（北北京京））股股份份有有限限公公司司16.【解析】（1）
……………………………………4分
（2）
…………………………………………9分
（3）
……………………………………………………………………………………11分
…………………………………………………………15分
(x−a) 2 +(y−b) 2 =r2
17.【解析】（方法一）（1）设圆的方程为 ，········································1分
A(0,−1) B(2,1)
因为圆C经过 和 ，且圆心在直线 上，
{(0−a) 2 +(−1−b) 2 =r2
(2−a) 2 +(1−b) 2 =r2
a−2b−1=0
,
所以 ·················································································································3分
高二数学试卷（第8页 共16页）
学学科科网网（（北北京京））股股份份有有限限公公司司{
a=1
b=0
r=√2
解得：
，········································································································································5分
所以圆C的方程为： ．·································································································6分
∵|CM|=|CN|=√2 90∘ | MN |= √ | CM | 2 +| CN | 2 =2
（2） ,且 = ∴弦长 ，·························7分
①当l斜率不存在时，l的方程为 ，
易知此时被圆C截得的弦长为2，符合题意．··················································································9分
②当l斜率存在时，设l的方程为 ，即 ，···························10分
又直线l被圆C所截得的弦长为2，所以 ，则 ．··················11分
所以 ，解得 ，······································································································13分
所以直线l的方程为 ，即 ．···························································14分
综上，l的方程为 或 ．························································································15分
（方法二）（1）依题意，设圆心 的坐标为 ，························································1分
则圆 的半径 ，··················································································································2分
所以 ，解得 ．··························································3分
所以圆心 ， ，·································································································4分
所以圆 的方程为 ．·····································································································6分
（2）因为 ，所以圆心 到直线 的距离 ，··········································8分
依题意，直线 的斜率不为0，设 的方程为 ，····················································9分
高二数学试卷（第9页 共16页）
学学科科网网（（北北京京））股股份份有有限限公公司司所以 ，································································································································12分
解得 ，或 ．································································································································14分
所以l的方程为 ，或 ，
即 或 ·························································································································15分
（方法三）（1）依题意， 的中点坐标为 ，·····································································1分
直线 的斜率 ，··············································································································2分
所以线段 的垂直平分线必过圆心 ，其方程为 ．
由 得 即 ，·······························································································3分
所以半径 ，········································································································4分
所以圆 的方程为 ．·····································································································6分
（2）同方法一．········································································································································12分
x2 +y2 +Dx+Ey+F=0
（方法四）（1）设圆的方程为 ，·························································1分
A(0,−1) B(2,1)
已知圆C经过 和 ，且圆心在直线 上.
{ (−1) 2 −E+F=0
22 +12 +2D+E+F=0
,
D E
(− )−2(− )−1=0
2 2
则 即 ··········································································3分
{D=−2
E=0
F=−1
解得：
，········································································································································5分
高二数学试卷（第10页 共16页）
学学科科网网（（北北京京））股股份份有有限限公公司司所以圆C的方程为 ．·······························································································6分
∵|CM|=|CN|=√2 90∘ | MN |= √ | CM | 2 +| CN | 2 =2
（2） ,且 = ∴弦长 ，·························7分
依题意，直线 的斜率不为0，设 的方程为 ，····················································8分
由 得 ，···············································9分
设 ，则
， ，············11分
所以 ，·················································································13分
即 ，即 ，解得 ，或 ．···················································14分
所以l的方程为 ，或 ，
即 或 ·························································································································15分
18.【解析】（1）证明：在直角梯形 中， ，
所以四边形 为矩形，故 ， ， ， ，
因为 为 的中点，所以 ，
，·························································································1分
在 中， ，所以
所以 ，故 ，································································································2分
因为 ，故 ，即 ，·····························3分
高二数学试卷（第11页 共16页）
学学科科网网（（北北京京））股股份份有有限限公公司司因为 ，翻折后有 ，
因为 ， 平面 ，所以 平面 ，···································5分
因为 平面 ，所以 ，·························································································6分
因为 ， 平面 ，所以 平面 ．·································7分
（2）由（1）可得 平面 ， ，所以 两两相互垂直，
以 为坐标原点， 所在直线分别为 轴建立空间直角坐标系 ，
所以 ，····································································································8分
假设存在点 符合题意，可设 ，（ ）·····················································10分
因为 ，所以 ，
即 ，所以 ，················································································11分
因为 ，
设 是平面 的法向量，则
取 ，则 ，故 ，·····················································13分
因为 轴垂直于平面 ，所以 是平面 的一个法向量，·····················14分
设平面 与平面 所成角为 ，则
，
解得 或 （舍去）··················································································································16分
所以 ，即 ，
高二数学试卷（第12页 共16页）
学学科科网网（（北北京京））股股份份有有限限公公司司故当点 位于靠近点 的三等分点时有平面 与平面 的夹角的余弦值为 .
···········································································································································································17分
19.【解析】（方法一）（1）设 ，则
直线 的斜率 ，直线 ，···················································2分
1
−
4
因为直线 和直线 的斜率乘积为 ，
y y 1
. =−
所以
x−2 x+2 4
，······························································································································3分
x2
+y2 =1 (x≠±2)
整理E的方程为 4 ．···································································································4分
（2）(ⅰ)依题意，设直线 的方程为 （ ）．
由 得 ， ·································································5分
A(x , y ), B(x , y )
设 1 1 2 2 ，则 ，即 ，
32k2 64k2 −4
x +x = , x x =
1 2 1+4k2 1 2 1+4k2
···············································································································6分
··························································································7分
．·································································································································8分
高二数学试卷（第13页 共16页）
学学科科网网（（北北京京））股股份份有有限限公公司司因为 ，所以 ，
所以 的取值范围为 ．··································································································10分
(ⅱ)直线 的方程为 ，直线 的方程为 ，·················11分
令 得 ，即 ．····················12分
所以 ， ·············································14分
因为 ，
所以 ，即 ， ···············································································16分
所以 ，
即 的值为1．·····································································································································17分
（方法二）（1）同方法一．···················································································································4分
（2）(ⅰ)依题意，直线 的斜率不为0，设 的方程为 ．
由 得 ，····················································································5分
A(x , y ), B(x , y )
设 1 1 2 2 ，则 ，解得 ，
，··········································································································6分
高二数学试卷（第14页 共16页）
学学科科网网（（北北京京））股股份份有有限限公公司司·································································································7分
．·································································································································8分
因为 ，所以 ，
所以 的取值范围为 ．··································································································10分
(ⅱ)直线 的方程为 ，直线 的方程为 ，·················11分
令 得 ， ，····························································12分
所以 ·························································································15分
．····················································································16分
所以 ，即 的值为1．·······································································································17分
（方法三）（1）同方法二．···················································································································4分
（2）(ⅰ)同方法二．·································································································································10分
高二数学试卷（第15页 共16页）
学学科科网网（（北北京京））股股份份有有限限公公司司(ⅱ)因为 ，所以 ，即 ，···········································12分
所以直线 的斜率 ，
同理可得 的斜率 ，······································································13分
所以 ，·········································································································15分
所以
．····················································································16分
所以 ，即 的值为1．·······································································································17分
高二数学试卷（第16页 共16页）
学学科科网网（（北北京京））股股份份有有限限公公司司
