文档内容
母题突破 3 定值问题
母题 已知抛物线C：y2＝2px(p>0)经过点P(1,2)，过点Q(0,1)的直线l与抛物线C有两个不
同的交点A，B，且直线PA交y轴于点M，直线PB交y轴于点N.
(1)求直线l斜率的取值范围；
(2)设O为原点，QM＝λQO，QN＝μQO，求证：＋为定值．
思路分析
❶联立l，C的方程，由判别式及PA，PB与y轴有交点求斜率的取值范围
↓
❷用A，B的坐标表示M，N的坐标
↓
❸用M，N的坐标表示λ，μ
↓
❹利用根与系数的关系计算＋
↓
❺求出＋为定值
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[子题1] (2022·张家口质检)已知双曲线C：x2－＝1，若直线l与双曲线C交于A，B两点，
且OA·OB＝0，O为坐标原点，证明：点O到直线l的距离为定值，并求出这个定值．
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[子题2] (2022·马鞍山模拟)已知椭圆C：＋＝1，P为椭圆上一点，过点P作斜率互为相反
数的两条直线，分别交椭圆于A，B两点(不与P点重合)，证明：直线AB的斜率为定值．
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规律方法 求解定值问题的两大途径(1)由特例得出一个值(此值一般就是定值)→证明定值：将问题转化为证明待证式与参数(某
些变量)无关．
(2)先将式子用动点坐标或动线中的参数表示，再利用其满足的约束条件使其绝对值相等的
正负项抵消或分子、分母约分得定值．
1．已知椭圆C：＋＝1的左、右焦点分别为F ，F ，点G是椭圆上一点．直线l：y＝kx＋m
1 2
与椭圆C交于A，B两点，且四边形OAGB为平行四边形．求证：平行四边形OAGB的面积
为定值．
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2．已知双曲线Γ：x2－＝1的左、右顶点分别为A(－1,0)，A(1,0)，过点F(2,0)斜率不为0
1 2
的直线l与Γ交于P，Q两点．记直线AP，AQ的斜率分别为k，k，求证：为定值．
1 2 1 2
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