2020年高考数学真题一题多解-526互联

(2020理科数学20)已知\(A,B\)为椭圆\(E:\dfrac{x^2}{a^2}+y^2=1(a>1)\)的左右顶点,\(G\)为\(E\)上的上顶点,\(\overrightarrow{AG}\cdot\overrightarrow{GB}=8,P\)为直线\(x=6\)上的动点,\(PA\)与\(E\)的另一个交点为\(C\),\(PB\)与\(C\)的另一交点为\(D\).

(1)求\(E\)的方程

(2)证明:\(CD\)过定点.

解:

(1).由题意得\(A(-a,0),B(a,0),G(0,1)\)则\(\overrightarrow{AG}=(a,1),\overrightarrow{GB}=(a,-1).\)由\(\overrightarrow{AG}\cdot\overrightarrow{GB}=8\)得\(a^2-1=8\)得\(a=3\).所以\(E\)的方程为\(\dfrac{x^2}{9}+y^2=1\)

(2). 题目的背景:设\(C(x_1,y_1),D(x_2,y_2),P(6,t)\)
我们首先观察下图形,挖深一下这个题的隐藏条件.题给了\(A,B\)两点和椭圆上的两点\(C,D\).我们不难想到连接\(AD,BD\)先利用一下椭圆的第三定义可以得到\(k_{AD}\cdot k_{BD}=-\dfrac{1}{9}\).再用一下题给条件,写出能够了解的数据\(k_{AP}=\dfrac{t}{9},k_{BD}=\dfrac{t}{3}\).则我们可以得到\(k_{AD}\cdot k_{AC}=-\dfrac{1}{9k_{BD}}=\dfrac{t}{9}\cdot\left(-\dfrac{1}{9}\right)\cdot \dfrac{3}{t}=-\dfrac{1}{27}\).而我们得到了这个隐藏的条件,从我们之前所讨论的,得到了\(k_{AD}\cdot k_{AC}\)为定值,那么\(DC\)一定是经过定点的. 因此此题是把一个常见模型给隐藏了.

法一:常规解法

设\(C(x_1,y_1),D(x_2,y_2),P(6,t)\)

若\(t\neq 0,\)设直线\(CD\)的方程为\(x=my+n\),由题意可知\(-3<n<3\)

由于直线\(PA\)的方程为\(y=\dfrac{t}{9}(x+3)\)
所以\(y_1=\dfrac{t}{9}(x_1+3)\)

直线\(PB\)的方程为\(y=\dfrac{t}{3}(x-3)\)
所以\(y_2=\dfrac{t}{3}(x_2-3)\)

可得\(3y_1(x-3)=y_2(x_1+3)\)

由于\(\dfrac{x_2^2}{9}+y_2^2=1\),故\(y_2^2=-\dfrac{(x_2+3)(x_2-3)}{9}\)

可得\(27y_1y_2=-(x_1+3)(x_2+3)\)

即

\[(27+m^2)y_1y_2+m(n+3)(y_1+y_2)+(n+3)^2=0(1.1) \]

将\(x=my+n\)与椭圆联立得

\[(m^2+9)y^2+2mny+n^2-9=0 \]

所以\(y_1+y_2=-\dfrac{2mn}{m^2+9},y_1y_2=\dfrac{n^2-9}{m^2+9}\)代回(1.1)得

\[(27+m^2)(n^2-9)-2m(n+3)mn+(n+3)^2(m^2+9)=0 \]

解得\(n=-3\)(舍)或\(n=\dfrac{3}{2}\)

故直线\(CD\)的方程为\(x=my+\dfrac{3}{2}\)

即直线\(CD\)过定点\(\left(\dfrac{3}{2},0\right)\)

若\(t=0,\)则直线\(CD\)为\(y=0\)肯定过点\(\left(\dfrac{3}{2},0\right)\)

综上\(CD\)过点\(\left(\dfrac{3}{2},0\right)\).

法二:搞硬的,算出\(CD\)方程

将\(PA\)与椭圆联立得到\((9+t^2)x^2+6t^2x+9t^2-81=0\)

则由韦达定理得到\(-3x_1=\dfrac{9t^2-81}{9+t^2}\)

从而\(x_1=\dfrac{-3t^2+27}{9+t^2},y_1=\dfrac{6t}{9+t^2}\),故\(C\left(\dfrac{-3t^2+27}{9+t^2},\dfrac{6t}{9+t^2}\right)\)

同理将\(PB\)与椭圆联立,一样得到\(D\left(\dfrac{3t^2-3}{1+t^2},\dfrac{-2t}{1+t^2}\right)\)

若直线\(CD\)斜率存在,则

\[k_{CD}=\dfrac{\dfrac{6t}{9+t^2}-\dfrac{-2t}{1+t^2}}{\dfrac{-3t^2+27}{9+t^2}-\dfrac{3t^2-3}{1+t^2}}=\dfrac{4t}{3(3-t^2)} \]

从而直线\(CD\)方程为

\[y=\dfrac{4t}{3(3-t^2)}\left(x-\dfrac{3t^2-3}{1+t^2}\right)-\dfrac{2t}{1+t^2}=\dfrac{4t}{3(3-t^2)\left(x-\dfrac{3}{2}\right)} \]

从而过定点\(\left(\dfrac{3}{2},0\right)\)

法三:齐次化处理

由之前讨论的\(k_{AC}\cdot k_{AD}=-\dfrac{1}{27}.\)

考虑坐标平移

\[(x,y)\rightarrow (x+3,y) \]

记为

\[(x,y)\rightarrow(x^{\prime},y^{\prime}) \]

\(x=x^{\prime}-3,y=y^{\prime}\),椭圆\(E\)可改写为

\[\dfrac{(x^{\prime}-3)^2}{9}+{y^{\prime}}^2=1\Rightarrow (1-6m){x^{\prime}}^2-6n{y^{\prime}}{x^{\prime}}+9{y^{\prime}}^2=0(1.2) \]

设平移后的直线\(CD\)方程为\(m{x^{\prime}}+n{y^{\prime}}=1\)代入(1.2)有

\[(1-6m){x^{\prime}}^2-6n{x^{\prime}}{y^{\prime}}+9{y^{\prime}}^2=0 \]

两边同除\({x^{\prime}}^2\)得

\[(1-6m)-6n{x^{\prime}}\dfrac{{y^{\prime}}}{{x^{\prime}}}+9\left(\dfrac{{y^{\prime}}}{{x^{\prime}}}\right)^2=0 \]

从而\(k_{AD}\cdot k_{CD}=\dfrac{1-6m}{9}=-\dfrac{1}{27}\)

解得\(m=\dfrac{2}{9}\)

从而平移后直线\(CD:\dfrac{2}{9}x+ny=1\)过定点\(\left(\dfrac{9 }{2},0\right)\)

将定点平移回原来的坐标轴得到\(CD\)过定点\(\left(\dfrac{3}{2},0\right)\)

法四:双联立

直线\(AC:y=\dfrac{t}{9}(x+3),BD:y=\dfrac{t}{3}(x-3)\)

两直线做乘得到过两直线所有的点的方程

\[27y^2-9ytx+27yt-3ytx-9yt+t^2(x^2-9)=0 \]

再联立椭圆\(\dfrac{x^2}{9}+y^2=1\)即\(x^2=9-9y^2\)
得

\[27y^2-12ytx+18yt-9y^2t^2=0 \]

消去\(y\)得\(CD\)直线

\[27y-12tx+18t-9yt^2=0 \]

直线过定点 \(\left(\dfrac{3 }{2},0\right)\)