2023-2024学年云南省高二期末模拟考试卷(24-579B)试题(数学)

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2023-2024学年云南省高二期末模拟考试卷(24-579B)试题(数学)试卷答案

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分析设A、B点坐标分别为(x1,y1),(x2,y2),由直线垂直的性质得x1x2+y1y2=0,由$\left\{\begin{array}{l}{x+2y-3=0}\\{{x}^{2}+{y}^{2}+x-2cy+c=0}\end{array}\right.$,得5y2-(2c+14)y+c+12=0,由此利用韦达定理能求出实数c的值.

解答解:设A、B点坐标分别为(x1,y1),(x2,y2),
由OA⊥OB,得kOA•kOB=-1,即$\frac{{y}_{1}}{{x}_{1}}•\frac{{y}_{2}}{{x}_{2}}$=-1,
∴x1x2+y1y2=0,①
由$\left\{\begin{array}{l}{x+2y-3=0}\\{{x}^{2}+{y}^{2}+x-2cy+c=0}\end{array}\right.$,
得5y2-(2c+14)y+c+12=0,
则${y}_{1}+{y}_{2}=\frac{1}{5}(2c+14)y+c+12=0$,②
又x1x2=(3-2y1)(3-2y2)=9-6(y1+y2)+4y1y2
代入①,得9-6(y1+y2)+5y1y2=0,③
联立②③,解得c=3.
∴实数c的值为3.

点评本题考查直线中参数值的求法,是中档题,解题时要认真审题,注意直线垂直的性质、韦达定理的合理运用.