非对称韦达定理的五种处理方法

$已知椭圆E:\cfrac{x^2}{a^2}+ \cfrac{y^2}{b^2}=1\quad (a\gt b\gt 0)的右顶点分别为A,B,离心率为\cfrac{\sqrt{3} }{2},$
$过P点(1,0)作直线交椭圆于C,D(与AB均不重合)，当点D与椭圆E的上顶点重合时，\left | AD \right |=\sqrt{5} $
$①求椭圆方程；$
$②设直线AD,BC的斜率分别为k_1,k_2,求证\cfrac{k_1}{k_2} 为定值。$
$解:设C(x_1,y_1),D(x_2,y_2),k_1=\cfrac{y_1}{x_1+2},k_2=\cfrac{y_2}{x_2-2}$
$\cfrac{k_1}{k_2} =\cfrac{y_1(x_2-2)}{y_2(x_1+2)}$
$反设直线CD方程:x=my+1$
$\begin{cases}x=my+1\\ x^2+4y^2-4=0 \end{cases}{\color{Green} \Rightarrow (m^2+4)y^2+2my-3=0} $
${\color{Red} y_1+y_2=\cfrac{-2m}{m^2+4} \quad y_1y_2=\cfrac{-3}{m^2+4} } $
$\cfrac{k_1}{k_2} =\cfrac{y_1(my_2+1-2)}{y_2(my_1+1+2)}=\cfrac{my_1y_2-y_1}{my_1y_2+3y_2}$
上式便是非对称韦达定理，处理方法有如下几种：
第一种：和积互化
${\color{Red} \cfrac{y_1+y_2}{y_1y_2} =\cfrac{2m}{3} }\Rightarrow my_1y_2=\cfrac{3}{2}(y_1+y_2) $
${\color{Red} \therefore \quad \cfrac{k_1}{k_2} =} \cfrac{my_1y_2-y_1}{my_1y_2+3y_2}=\cfrac{{\color{Red} \cfrac{3}{2}(y_1+y_2)} -y_1}{{\color{Red} \cfrac{3}{2}(y_1+y_2)} +3y_1}= \cfrac{1}{3} $
和积互化还可参见：https://one.free.nf/index.php/archives/3/ 中例二法三
第二种方法：用曲线方程替换

$x_1^2+4y_1^2=4\Rightarrow x_1^2=4-4y_1^2\Rightarrow (x_1+2)(x_1-2)=-4y_1^2$

$\Rightarrow(x_1+2)=\cfrac{-4y_1^2}{(x_1-2)} 代入\cfrac{k_1}{k_2} =\cfrac{y_1(x_2-2)}{y_2(x_1+2)}$
第三种暴力硬解法：
${\color{Green}(m^2+4)y^2+2my-3=0} \Rightarrow y=\cfrac{-2m\pm \sqrt{4m^2+12(m^2+4)} }{2(m^2+4)} $
${\color{Red} =\cfrac{-m\pm 2\sqrt{m^2+3} }{m^2+4} } $
$\cfrac{k_1}{k_2} =\cfrac{y_1(x_2-2)}{y_2(x_1+2)}＝\cfrac{my_1y_2-y_1}{my_1y_2+3y_2}
=\cfrac{m\cdot \cfrac{-3}{m^2+4} -{\color{Red} \cfrac{-m- 2\sqrt{m^2+3} }{m^2+4} }}{m\cdot\cfrac{-3}{m^2+4} +3\cdot {\color{Red} \cfrac{-m+ 2\sqrt{m^2+3} }{m^2+4} }}
$
$\cfrac{-3m {\color{Red} +m+ 2\sqrt{m^2+3} }}{-3m{\color{Red} -3m+ 6\sqrt{m^2+3 } }}=\cfrac{1}{3} $
第四种方法：利用圆锥曲线第三定义，前提是两定点是顶点
https://one.free.nf/index.php/archives/3/
例2中法二，如果同时齐次化计算会更简化。
齐次化操作见：https://one.free.nf/index.php/category/%E5%9C%86%E9%94%A5/1/

第五种方法：保留一个未知数，再配凑
$\cfrac{k_1}{k_2} =\cfrac{y_1(x_2-2)}{y_2(x_1+2)}＝\cfrac{my_1y_2-y_1}{my_1y_2+3y_2}=\cfrac{my_1y_2-{\color{Red} (y_1+y_2)-y_2} }{my_1y_2+3y_2}$
$=\cfrac{\cfrac{-3m}{m^2+4}-\cfrac{-2m}{m^2+4}+y_2}{\cfrac{-3m}{m^2+4}+3y_2 }=\cfrac{\cfrac{-m}{m^2+4}+y_2}{\cfrac{-3m}{m^2+4}+3y_2 } =\cfrac{1}{3} $