2024学生版大二轮数学新高考提高版（京津琼鲁辽粤冀鄂湘渝闽苏浙黑吉晋皖云豫新甘贵赣桂）专题六　第4讲　母题突破3　定值问题31

这是一份2024学生版大二轮数学新高考提高版（京津琼鲁辽粤冀鄂湘渝闽苏浙黑吉晋皖云豫新甘贵赣桂）专题六　第4讲　母题突破3　定值问题31，共3页。


(1)求椭圆E的方程；
(2)如图，过点B(0,4)作一条与y轴不重合的直线，该直线交椭圆E于C，D两点，直线AD，AC分别交x轴于H，G两点，O为坐标原点．证明：|OH||OG|为定值，并求出该定值．
思路分析
❶结合点的坐标和AM的斜率列方程组
❷设直线BC的方程并与椭圆的方程联立
❸得到x1＋x2，x1x2
❹写出直线AD，AC的方程并求出H，G的横坐标
❺化简运算|OH||OG|
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[子题1] (2023·西安模拟)如图，在平面直角坐标系中，椭圆E：eq \f(x2,8)＋eq \f(y2,4)＝1，A，B分别为椭圆E的左、右顶点．已知图中四边形ABCD是矩形，且|BC|＝4，点M，N分别在边BC，CD上，AM与BN相交于第一象限内的点P.若点P在椭圆E上，证明：eq \f(|BM|,|CN|)为定值，并求出该定值．
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[子题2] (2023·衡水质检)已知E(2,2)是抛物线C：y2＝2x上一点，经过点(2,0)的直线l与抛物线C交于A，B两点(不同于点E)，直线EA，EB分别交直线x＝－2于点M，N.已知O为原点，求证：∠MON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．已知抛物线C：y2＝2px(p>0)，F为其焦点，若圆E：(x－1)2＋y2＝16与抛物线C交于A，B两点，且|AB|＝4eq \r(3).
(1)求抛物线C的方程；
(2)若点P为圆E上任意一点，且过点P可以作抛物线C的两条切线PM，PN，切点分别为M，N.求证：|MF|·|NF|恒为定值．
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2．(2023·沧州模拟)已知双曲线C：eq \f(x2,a2)－eq \f(y2,b2)＝1(a>0，b>0)经过点eq \b\lc\(\rc\)(\a\vs4\al\c1(3，\f(\r(6),2)))，右焦点为F(c,0)，且c2，a2，b2成等差数列．
(1)求C的方程；
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(2)过F的直线与C的右支交于P，Q两点(P在Q的上方)，PQ的中点为M，M在直线l：x＝2上的射影为N，O为坐标原点，设△POQ的面积为S，直线PN，QN的斜率分别为k1，k2，证明：eq \f(k1－k2,S)是定值．
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