2026届高三毕业班教学质量检测数学参考答案及评分细则第1页（共9页）2026届高三毕业班教学质量检测数学参考答案及评分细则评分说明：1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制定相应的评分细则。2．对计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分。3．解答右端所注分数，表示考生正确做到这一步应得的累加分数。4．只给整数分数。选择题和填空题不给中间分。一、单项选择题：本大题共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．D2．B3．A4．A5．D6．A7．C8．C二、多项选择题：本大题共3小题，每小题6分，共18分。在每小题给出的四个选项中，有多个选项符合题目要求，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．ABC10．ABC11．ACD三、填空题：本大题共3题，每小题5分，共15分。12．7π13．01yx14．50四、解答题：本大题共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．解法一：（1）由题意，得1222222||4,461,9,0,cFFabcabab··················································3分解得6,2,2,abc·······················································································5分所以C的方程为22162xy．·································································6分（2）设00(,)Qxy，00x，00y，由题意，得1(2,0)F，2(2,0)F．因为点Q在C上且在第一象限，12QFF△为直角三角形，所以2190QFF，或1290FQF．···················································7分若2190QFF，则6(2,)3Q；····························································8分若1290FQF，则2212000000(2,)(2,)40QFQFxyxyxy，{#{QQABSQC9xgCQkJbACR4LQ0HIC0mYsJEgJMgMBRCYqARDyQFABIA=}#}2026届高三毕业班教学质量检测数学参考答案及评分细则第2页（共9页）·········································································································11分又Q在C上，所以2200162xy，·····························································12分所以003,1,xy即(3,1)Q．综上，点Q的坐标为6(2,)3，或(3,1)．················································13分解法二：（1）由题意，得椭圆C的左、右焦点分别为12(2,0),(2,0)FF，点6(2,)3P在C上，······························································································1分所以2c，12662||||162639aPFPF，·····························3分所以6a，222bac，··························································5分所以C的方程为22162xy．······························...