2024學生版大二輪數學新高考提高版（京津瓊魯遼粵冀鄂湘渝閩蘇浙黑吉晉皖云豫新甘貴贛桂）專題六　培優(yōu)點8　圓錐曲線中非對稱韋達定理的應用27

培優(yōu)點8圓錐曲線中非對稱韋達定理的應用在圓錐曲線問題中，我們聯立直線和圓錐曲線方程，消去x或y，得到一個一元二次方程，往往能夠利用韋達定理來快速處理|x1－x2|，xeq\o\al(2,1)＋xeq\o\al(2,2)，eq\f(1,x1)＋eq\f(1,x2)之類的結構，但在有些問題中，我們會遇到涉及x1，x2的不同系數的代數式的運算，比如求eq\f(x1,x2)，eq\f(3x1x2＋2x1－x2,2x1x2－x1＋x2)或λx1＋μx2之類的結構，我們把這種系數不對等的結構，稱為“非對稱韋達結構”．考點一分式型例1(2023·新高考全國Ⅱ)已知雙曲線C的中心為坐標原點，左焦點為(－2eq\r(5)，0)，離心率為eq\r(5).(1)求C的方程；(2)記C的左、右頂點分別為A1，A2，過點(－4,0)的直線與C的左支交于M，N兩點，M在第二象限，直線MA1與NA2交于點P.證明：點P在定直線上．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________規(guī)律方法非對稱結構的常規(guī)處理方法有和積轉換、配湊、求根公式(暴力法)、曲線方程代換、第三定義等方法，將其轉化為對稱結構計算．跟蹤演練1已知橢圓C：eq\f(x2,a2)＋eq\f(y2,b2)＝1(a>b>0)的左、右焦點分別為F1(－c,0)，F2(c,0)，M，N分別為左、右頂點，直線l：x＝ty＋1與橢圓C交于A，B兩點，當t＝－eq\f(\r(3),3)時，A是橢圓的上頂點，且△AF1F2的周長為6.(1)求橢圓C的方程；(2)設直線AM，BN交于點Q，證明：點Q在定直線上．(3)設直線AM，BN的斜率分別為k1，k2，證明：eq\f(k1,k2)為定值．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________考點二比值型例2(2023·深圳模擬)在平面直角坐標系Oxy中，已知雙曲線C：eq\f(y2,a2)－eq\f(x2,b2)＝1(a>0，b>0)的一條漸近線為y＝eq\f(\r(3),3)x，且點Peq\b\lc\(\rc\)(\a\vs4\al\co1(\r(3)，\r(2)))在C上．(1)求C的方程；(2)設C的上焦點為F，過F的直線l交C于A，B兩點，且eq\o(AF,\s\up6(→))＝7eq\o(BF,\s\up6(→))，求l的斜率．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________規(guī)律方法比值型問題適用于x1＝λx2型，可以采用倒數相加，但有時得到的可能不是這種形式，而是x1＝λx2＋k的形式，此時采用待定系數法，例如x1＝－3x2＋4，可以轉化x1－1＝－3(x2－1)，得到eq\f(x1－1,x2－1)＝－3，繼續(xù)采用倒數相加解決．跟蹤演練2已知點A(0，－2)，橢圓E：eq\f(x2,a2)＋eq\f(y2,b2)＝1(a>b>0)的離心率為eq\f(\r(3),2)，F是橢圓E的右焦點，直線AF的斜率為eq\f(2\r(3),3)，O為坐標原點．(1)求E的方程；(2)設過點A的直線l與E相交于P，Q兩點，且eq\o(AP,\s\up6(→))＝eq\f(1,2)eq\o(AQ,\s\up6(→))，求△OPQ的面積及直線l的方程．____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________