$若函数f(x)=(1-x^2)(x^2+ax+b)的图像关于直线x=3对称，则f(x)的最大值为(\qquad )$
$A.16,\quad B.10+8\sqrt{10}\quad C.36\quad D.19+6\sqrt{10}$

${\color{Red} 弦的斜率与弦中点与原点连线的斜率之积为定值e^2-1} $
$例1、2017年新课标1之20题，不过椭圆\cfrac{x^2}{4}+y^2=1的点P(0,1) 的直线l交椭圆于A,B两点，若直线PA,PB的斜率之和为-1，证明l过定点。$
斜率双用前置知识一，${\color{Red}椭圆上两点 } (x_1,y_1)(x_2,y_2),$用点差法可得

$\begin{cases} \cfrac{x^2_1}{a^2} +\cfrac{y^2_1}{b^2}=1 \quad \\ \quad \\\cfrac{x^2_2}{a^2} +\cfrac{y^2_2}{b^2}=1 \end{cases}$
两式相差，得$\cfrac{y_1^2-y_2^2}{x_1^2-x_2^2} =-\cfrac{b^2}{a^2} =e^2-1$
${\color{Red} \Rightarrow \cfrac{y_1-y_2}{x_1-x_2}=(e^2-1)\cdot \cfrac{x_1+x_2}{y_1+y_2} } $

斜率双用前置知识二两点式的直线方程：$\cfrac{y_1-y}{x_1-x}=\cfrac{y_1-y_2}{x_1-x_2}\Rightarrow {\color{Red} x_1y_2-x_2y_1= x(y_2-y_1)+y(x_1-x_2)}$

$P(0,1)设A点坐标为(x_1,y_1),B(x_2,y_2),{\color{Red} PAB均在椭圆上，}$
${\color{Green} 由Ｐ(0,1),A(x_1,y_1)点差有：}k_{PA}=\cfrac{y_1-1}{x_1} =(e^2-1)\cdot \cfrac{x_1}{y_1+1}$
${\color{Green} 由Ｐ(0,1),B(x_2,y_2)点差有：}k_{PB}=\cfrac{y_2-1}{x_2} =(e^2-1)\cdot \cfrac{x_2}{y_2+1},即斜率的两种表示方式$
将斜率的两种表示“轮换”代入已知条件，即$k_{PA}+k_{PB}=-1$

$\begin{cases} \cfrac{y_1-1}{x_1}-\cfrac{1}{4}\cdot \cfrac{x_2}{y_2+1}=-1\\\quad \\ \cfrac{y_2-1}{x_2}-\cfrac{1}{4}\cdot \cfrac{x_1}{y_1+1}=-1 \end{cases}$

这里轮换的意思是，第一个斜率用直线表示，第二个斜率用椭圆表示。

化为整式得，$\begin{cases} 4(y_1y_2+y_1-y_2-1)-x_1x_2=-4x_1y_2-4x_1\\\quad \\ 4(y_1y_2+y_2-y_1-1)-x_1x_2=-4x_2y_1-4x_2 \end{cases}$
两式相减速，得$x_2y_1-x_1y_2=2(y_1-y_2)-(x_2-x_1)$
对比两点式的直线方程，${\color{Red} x_1y_2-x_2y_1= x(y_2-y_1)+y(x_1-x_2)},得直线过定点(2,-1)$

$例2、已知椭圆C:\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1(a\gt b\gt 0)的离心率为\cfrac{\sqrt{2} }{2} ,$
$以Ｃ的短轴为直径的圆与直线y=ax+6相切。$

$(1)求Ｃ的方程；$
$(2)直线l:y=K(x-1)(k\ge 0)与Ｃ相交于A,B两点，过Ｃ上的点Ｐ作x轴的平行线相交线AB于点Ｑ，$
$直线OP的斜率为{k}' (O为原点),\triangle APQ的面积为S_1.\triangle BPQ的面积为S_2.$
$若\left | AP \right | \cdot S_2=\left | BP \right | \cdot S_1,判断k\cdot {k}'是否为定值？并说明理由。$

解：$(1)\begin{cases} e=\cfrac{c}{a} =2\sqrt{2}\\ \cfrac{\left | 6 \right | }{\sqrt{a^2+1} }=b\\a^2=b^2+c^2 \end{cases}\Rightarrow \begin{cases}a=2\sqrt{2}\\b=2\\c=2\end{cases}$
则椭圆Ｃ的方程为：$\cfrac{x^2}{8}+\cfrac{y^2}{4}=1$
$(2)\cfrac{\left | AP \right | }{\left | BP \right | } =\cfrac{S_1}{S_2}=\cfrac{\frac{1}{2} \left | AP \right |\left | PQ \right |\sin\angle APQ}{\frac{1}{2} \left | BP \right |\left | PQ \right |\sin\angle BPQ}{\color{Red} \Rightarrow \sin\angle APQ=\sin\angle BPQ}$
而$\angle APQ+\angle BPQ=\angle APB\in(0,\pi),则\angle APQ=\angle BPQ\Rightarrow k_{PA}+k_{PB}=0$
${\color{Red}PAB三点在椭圆上 },设P(x_0,y_0),A(x_1,y_1),B(x_2,y_3),则有：$
${\color{Green} 由P(x_0,y_0),A(x_1,y_1)点差有：}k_{PA}：\cfrac{y_1-y_0}{x_1-x_0} =-\cfrac{1}{2} \cdot \cfrac{x_1+x_0}{y_1+y_0}$
${\color{Green} 由P(x_0,y_0),B(x_2,y_3)点差有：}k_{PB}：\cfrac{y_2-y_0}{x_2-x_0} =-\cfrac{1}{2} \cdot\cfrac{x_2+x_0}{y_2+y_0}$
$\Rightarrow \begin{cases} \cfrac{y_1-y_0}{x_1-x_0}-\cfrac{1}{2}\cdot \cfrac{x_2+x_0}{y_2+y_0}=0\\ \quad \\ \cfrac{y_2-y_0}{x_2-x_0} -\cfrac{1}{2} \cdot\cfrac{x_1+x_0}{y_1+y_0} =0\end{cases}$
$\Rightarrow \begin{cases}2(y_1-y_0)(y_2+y_0)=(x_2+x_0)(x_1-x_0)\\ \quad \\2(y_2-y_0)(y_1+y_0)=(x_1+x_0)(x_2-x_0) \end{cases}$
$\Rightarrow \begin{cases}2y_1y_2+2y_0y_1-2y_0y_2-2y_0^2=x_0x_1+x_1x_2-x_0x_2-x_0^2\\ \quad \\2y_1y_2+2y_0y_2-2y_0y_1-2y_0^2=x_0x_2+x_1x_2-x_0x_1-x_0^2 \end{cases}$
$两式相差，得2y_0(y_1-y_2)=x_0(x_1-x_2)$
$k\cdot {k}' =\cfrac{y_1-y_2}{x_1-x_2} \cdot \cfrac{y_0}{x_0}=\cfrac{1}{2} $
$说明：题目中没有直接告诉你k_{PA}+k_{PB}=0,这就是押轴题目的套路，拐一个弯而已$

$例3、双曲线C:\cfrac{x^2}{a^2}-\cfrac{y^2}{b^2}=1(a\gt b\gt b\gt 0)的左顶点为A,焦距为4,$
$过右焦点Ｆ作垂直于实轴的直线交Ｃ于B,D两点，且\triangle ABD是直角三角形。$
$(1)求双曲线C的方程；$
$(2)M、N是双曲线C右支上的两动点，设直线AM,AN的斜率分别为k_1,k_2,若k_1k_2=-2,$
$求点A到直线MN的距离d的取值范围。$

$(1)解：\begin{cases} 2c=4\\ \cfrac{b^2}{a}=a+c\\a^2+b^2=c^2 \end{cases}\Rightarrow \begin{cases} c=2,\\a=1, \\b=\sqrt{3}\end{cases}$
$\Rightarrow x^2-\cfrac{y^2}{3}=1$
$(2)A(-1,0),设M(x_1,y_1),N(x_2,y_2),$
因为${\color{Red} 弦的斜率与弦中点与原点连线的斜率之积为定值e^2-1}$
${\color{Green} 由A(-1,0),M(x_1,y_1)点差有},k_{AM}=k_1=\cfrac{y_1}{x_1+1}=\cfrac{b^2}{a^2}\cdot \cfrac{x_1-1}{y_1}$
${\color{Green} 由A(-1,0),N(x_2,y_2)点差有}, k_{AN}=k_2=\cfrac{y_2}{x_2+1}=\cfrac{b^2}{a^2}\cdot \cfrac{x_2-1}{y_2}$
$\Rightarrow \begin{cases} \cfrac{y_1}{x_1+1}\cdot\cfrac{3x_2-3}{y_2}=-2 \\\cfrac{y_2}{x_2+1}\cdot\cfrac{3x_1-3}{y_1}=-2 \end{cases}$
$\Rightarrow \begin{cases} 2y_2(x_1+1)+3y_1(x_2-1)=0\\\quad \\2y_1(x_2+1)+3y_2(x_1-1)=0\end{cases}\Rightarrow \begin{cases} 2x_1y_2+2y_2+3x_2y_1-3y_1=0 \\ \quad \\2x_2y_1+2y_1+3x_1y_2-3y_2=0 \end{cases}$
两式相减，得$x_1y_2-x_2y_1=5(y_2-y_1)$
对比直线两点式的变形：
${\color{Green} \cfrac{y_1-y_2}{x_1-x_2} =\cfrac{y_1-y}{x_1-x} \Rightarrow x_1y_2-x_2y_1=x(y_2-y_1)+y(x_1-x_2)}$
可知直线MN过定点$(5,0)$
直线MN过定$(5,0)仅与双曲线右支有两交点的斜率为渐近线内，即倾斜角\alpha \in (\cfrac{\pi}{3},\cfrac{2\pi}{3} )$
$d\in (6\sin \cfrac{\pi}{3} ,6]即d\in (3\sqrt{3} ,6]$

$例4、已知点A(2,1)在双曲线C:\cfrac{x^2}{a^2}+\cfrac{y^2}{a^2-1}=1(a\gt 1)$上，直线$l交C于P，Q两点,直线AP,AQ$的斜率之和为$0$.
$①求l的斜率。$
$②若\tan \angle PAQ=2\sqrt{2},求\triangle PAQ的面积$

$(1)依题意，有\cfrac{2^2}{a^2} -\cfrac{1}{a^2-1} =1\Rightarrow a=\sqrt{2}$
$双曲线C方程为：\cfrac{x^2}{2} -y^2=1$
$设P(x_1,y_1),Q(x_2,y_2),根据点差法有：$
$\begin{cases} \cfrac{y_1-1}{x_1-2}=\cfrac{x_1+2}{2(y_1+1)} \\ \quad \\ \cfrac{y_2-1}{x_2-2}=\cfrac{x_2+2}{2(y_2+1)}\end{cases}$
$k_{AP}+k_{AQ}\Rightarrow \begin{cases} \cfrac{y_1-1}{x_1-2}+\cfrac{x_2+2}{2(y_2+1)}=0 \\ \quad \\ \cfrac{y_2-1}{x_2-2}+\cfrac{x_1+2}{2(y_1+1)}=0 \end{cases}$
$\Rightarrow\begin{cases} y_1y_2+\frac{1}{2}x_1x_2 -(y_2-y_1)-(x_2-x_1)-3=0 \\ \quad \\ y_1y_2+\frac{1}{2}x_1x_2 -(y_1-y_2)-(x_1-x_2)-3=0 \end{cases}$
两式相减，并整理得：$y_2-y_1=-(x_2-x_1)\Rightarrow k_{PQ}=-1$
(2)$\begin{cases} \tan \angle PAQ=2\sqrt{2}\\ \quad \\ k_{AP}+k_{AQ}=0\end{cases}\Rightarrow \begin{cases} k_{AP}=-\sqrt{2} \\\quad \\k_{AQ}=\sqrt{2} \end{cases}$
设$\overrightarrow{AP}=\lambda (1,-\sqrt{2} )\Rightarrow P(2+\lambda,1-\sqrt{2} \lambda),代入曲线C得，\lambda=\cfrac{4+4\sqrt{2} }{3}$
设$\overrightarrow{AQ}=\mu (1,\sqrt{2} )\Rightarrow P(2+\mu ,1+\sqrt{2} \mu ),代入曲线C得，\mu =\cfrac{4-4\sqrt{2} }{3}$
$s_{\triangle PAQ}={\color{Red} \cfrac{1}{2}\overrightarrow{AP} \cdot \overrightarrow{AQ} \tan \angle PAQ} =\cfrac{1}{2}\lambda\mu (1,-\sqrt{2})(1,\sqrt{2})\tan \angle PAQ=\cfrac{16}{9}\sqrt{2}$

练习：
1、$椭圆C：\cfrac{x^2}{4}+\cfrac{y^2}{3}=1左顶点为A，不过A点斜率为k的直线与椭圆C交于M、N两点，$
$记AM,AN的斜率为k_1,k_2,k_1+k_2=\cfrac{3}{k},证明l过定点$
$解：A(-2,0),M(x_1,y_1),N(x_2,y_2)$
$点差法：\begin{cases}\cfrac{y_1}{x_1+2} \cdot \cfrac{y_1}{x_1-2}=-\cfrac{3}{4}\\ \cfrac{y_2}{x_2+2} \cdot \cfrac{y_2}{x_2-2}=-\cfrac{3}{4} \end{cases}\Rightarrow\begin{cases} \cfrac{y_1}{x_1+2}-\cfrac{3}{4}\cdot \cfrac{x_2-2}{y_2}=\cfrac{3}{k}\\ \cfrac{y_2}{x_2+2}-\cfrac{3}{4}\cdot \cfrac{x_1-2}{y_1}=\cfrac{3}{k}\end{cases}$
$两式相减，得x_1y_2-x_2y_1=k(x_1-x_2)-2(y_2-y_1)$
$对比直线两点式的变形，可得，直线l过点(-2,k)$
$y-k=k(x+2)\Rightarrow y=k(x+3),直线过定点(-3,0)$

$\vec{u} 和\vec{v}$为邻边张成的平行四边形，可以通过向量的叉积的大小来计算其面积。在二维中，叉积的大小实际上就是行列式的绝对值。
$\begin{vmatrix}x_1 & y_1\\x_2&y_2\end{vmatrix}=x_1y_2-x_2y_1$
$\left | \begin{vmatrix}x_1 & y_1\\x_2&y_2\end{vmatrix} \right | =\left | x_1y_2-x_2y_1 \right | $
证明法一：
从几何上，平行四边形的面积可以通过底乘高来计算。假设以$\vec{u}$为底，那么高是$\vec{v}$在垂直于的方向上的投影长度。$设\vec{u} =(a,c)和\vec{v}=(b,d)。平行四边形面积A为：$
$$A=\left | \vec{u} \right | \left | \vec{v} \right | \left | \sin \theta \right |$$

$底长=\sqrt{a^2+c^2} \cdot ，高是\vec{v}$在垂直于的方向上的投影长度
$垂直于\vec{u}的单位向量是\vec{n}=\cfrac{(-c,a)}{\sqrt{a^2+c^2} } ,所以高为\left | \vec{v}\cdot \vec{n}\right |= \left | \cfrac{-bc+ad}{\sqrt{a^2+c^2} } \right |$
$A= \left | \vec{u} \right | \left |\vec{v}\cdot \vec{n} \right | =\sqrt{a^2+c^2} \cdot \left | \frac{-bc+ad}{\sqrt{a^2+c^2} } \right |=\left | -bc+ad \right |$

法二：
$A^2=\left | \vec{u} \right | ^2 \left | \vec{v} \right | ^2-( \vec{u} \cdot \vec{v} )^2=(a^2+c^2)(b^2+d^2)-(ab+cd)^2=a^2d^2+b^2c^2-2abcd=(ad-bc)^2$
代入坐标即可。
法二也是证明平方和恒等式${\color{Green} (a^2+c^2)(b^2+d^2)=(ab+cd)^2+(ad-bc)^2} $的方法。
https://uu.890222.xyz/index.php/archives/260/
https://uu.890222.xyz/index.php/category/%E5%9C%86%E9%94%A5/2/ 例7

1.斜率之积为定值
大家还记得这道题的考点吗：$椭圆 \cfrac{x^2}{a^2}+ \cfrac{y^2}{b^2}=1 上存在三点A(x_0,y_0),M,N，且满足k_{AM}k_{AN}=\lambda ，$
$\lambda \ne\cfrac{a^2}{b^2},那么直线过定点P(tx_0,-ty_0)(t=\cfrac{a^2\lambda +b^2}{a^2\lambda -b^2})$。
证明方法，用平移齐恣法。
2.斜率之和为定值
下面让我们转换视角，看看斜率之和为定值的情况：
$椭圆\cfrac{x^2}{a^2}+ \cfrac{y^2}{b^2}=1 上存在三点A(x_0,y_0),M,N，且满足k_{AM}+k_{AN}=\lambda (\lambda \ne0)，$
$那么直线过定点P(x_0-\cfrac{2y_0}{\lambda},-y_0+\cfrac{2(e^2-1)x_0}{\lambda})=(x_0-\cfrac{2y_0}{\lambda},-y_0-\cfrac{2b^2x_0}{a^2\lambda})$。
$证明：记k_1=k_{AM},k_2=k_{AN}$
$作平移变换：\begin{cases} {x}'=x-x_0\\\quad\\{y}'=y-y_0 \end{cases}\Rightarrow \begin{cases} x={x}'+x_0\\\quad \\y={y}'+y_0 \end{cases}$
$使得A(x_0,y_0)成为新坐标系的原点{A}'(0,0)$
$\cfrac{x^2}{a^2}+ \cfrac{y^2}{b^2}=1 \Rightarrow \cfrac{({x}'+x_0)^2}{a^2}+ \cfrac{({y}'+y_0)^2}{b^2}=1$
$a^2{y}'^2+2(b^2x_0{x}'+a^2y_0{y}')+b^2{x}'^2=0$
$平移后的直线{M}'{N}':m{x}'+n{y}'=1,上式一次项乘上m{x}'+n{y}',$
$a^2{y}'^2+2(b^2x_0{x}'+a^2y_0{y}')(m{x}'+n{y}')+b^2{x}'^2=0$
$\Rightarrow a^2(1+2ny_0){y}'^2+2(nb^2x_0+ma^2y_0){x}'{y}'+b^2(1+2mx_0){x}'^2=0$
$除以{x}'^2,得a^2(1+2ny_0)\cfrac{{y}'^2}{{x}'^2}+2(nb^2x_0+ma^2y_0)\cfrac{{y}'}{{x}'}+b^2(1+2mx_0)=0$
$\Rightarrow a^2(1+2ny_0){k}'^2+2(nb^2x_0+ma^2y_0){k}'+b^2(1+2mx_0)=0$
$由韦达定理，得{k_1}' +{k_2}' =-\cfrac{2(nb^2x_0+ma^2y_0)}{a^2(1+2ny_0)},{k_1}' {k_2}' =\cfrac{b^2(1+2mx_0)}{a^2(1+2ny_0)}$

1.斜率之积为定值:
${k_1}' {k_2}' =\cfrac{b^2(1+2mx_0)}{a^2(1+2ny_0)}=\lambda:t=\cfrac{b^2}{a^2\lambda},$
$则：t(1+2mx_0)=1+2ny_0\Rightarrow n=\cfrac{t(1+2mx_0)-1}{2y_0}$
$所以，m{x}'+n{y}'=1\Rightarrow m{x}'\cfrac{t(1+2mx_0)-1}{2y_0}{y}'=1$
$因为 m是设出来的参数，而我们要找的是直线 {M}'{N}'所过的定点，也就是要找到关于$
${x}',{y}'的恒等式，那就要消掉参数m，也就是让和 m相乘的式子等于 0，$
所以整理式子得：$ m({x}'+\cfrac{x_0t}{y_0}{y}')+\cfrac{t-1}{2y_0}{y}'=1$，
$\begin{cases} \cfrac{t-1}{y_0}{y}'=0\\ \quad \\{x}'+\cfrac{x_0t}{y_0}{y}'=0\end{cases}$
$\Rightarrow \begin{cases} {y}'=\cfrac{2y_0}{t-1}\\ \quad \\{x}'=\cfrac{2x_0t}{1-t}\end{cases}再代入 \begin{cases} x={x}'+x_0\\ \quad \\y={y}'+y_0\end{cases}\Rightarrow $
$\begin{cases} x=\cfrac{2x_0t}{1-t}+x_0=\cfrac{t+1}{1-t}x_0=\cfrac{x_0(b^2+\lambda a^2)}{a^2\lambda -b^2}\\ \quad \\y=\cfrac{2y_0}{t-1}+y_0=\cfrac{t+1}{t-1}y_0=\cfrac{y_0(b^2+\lambda a^2)}{-(a^2\lambda -b^2)}\end{cases}\Rightarrow$

$也就是说直线MN过定点P(\cfrac{x_0(b^2+\lambda a^2)}{a^2\lambda -b^2},\cfrac{y_0(b^2+\lambda a^2)}{-(a^2\lambda -b^2)})$

2.斜率之和为定值$\lambda$
$斜率之和{k_1}' +{k_2}' =-\cfrac{2(nb^2x_0+ma^2y_0)}{a^2(1+2ny_0)}=\lambda:记t=\cfrac{b^2}{a^2}$
$则：nb^2x_2+ma^2y_0=\cfrac{\lambda a^2(1+2ny_0)}{-2}\Rightarrow m=\cfrac{\lambda(\cfrac{1}{2}+ny_0)+ntx_0}{-y_0},$
$m{x}'+n{y}'=1,所以,\Rightarrow \cfrac{\lambda(\cfrac{1}{2}+ny_0)+ntx_0}{-y_0}{x}'+n{y}'=1$
$因为 n是设出来的参数，而我们要找的是直线 {M}'{N}'所过的定点，也就是要找到关于$
${x}',{y}'的恒等式，那就要消掉参数n，也就是让和 n相乘的式子等于 0，$
所以，整理式子得：$n[{y}'-(\lambda+\cfrac{tx_0}{y_0}{x}')]-\cfrac{\lambda}{2y_0}{x}'=1$
$\begin{cases} {x}'=-\cfrac{2y_0}{\lambda } \\{y}'=-2(y_0+\cfrac{tx_0}{\lambda } ) \end{cases}$再代入$\begin{cases} x={x}'+x_0\\ \quad \\y={y}'+y_0\end{cases}\Rightarrow $
$\begin{cases} x=-\cfrac{2y_0}{\lambda } +x_0\\ \quad \\y=-2(y_0+\cfrac{tx_0}{\lambda } )+y_0\end{cases}$
也就是说直线MN过定点$P(x_0-\cfrac{2y_0}{\lambda},-y_0+\cfrac{2(e^2-1)x_0}{\lambda})=(x_0-\cfrac{2y_0}{\lambda},-y_0-\cfrac{2b^2x_0}{a^2\lambda})$

适用于圆锥曲线中的直线过轴点时使用。
比如：$例1中直线过T(4,0),例2中直线过T(2,0),例3中直线过T(0,4)。$
前置知识：$动直线l过定点(t,0)交椭圆\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1于两点(x_1,y_1),(x_2,y_2)$
根据直线的两点式有：$\cfrac{y_1}{x_1-t}=\cfrac{y_2}{x_2-t}\Rightarrow x_1y_2-x_2y_1=t(y_2-y_1)$
构造$x_1y_2-x_2y_1的对偶式x_1y_2+x_2y_1=\cfrac{(x_1y_2)^2-(x_2y_1)^2}{x_1y_2-x_2y_1}$
$=\cfrac{x_1^2y_2^2-x_2^2y_1^2}{t(y_2-y_1)}=\cfrac{a^2(1-\cfrac{y_1^2}{b^2})^2y_2^2-a^2(1-\cfrac{y_2^2}{b^2})^2y_1^2}{t(y_2-y_1)}=\cfrac{a^2(y_2^2-y_1^2)}{t(y_2-y_1)}=\cfrac{a^2}{t}(y_2+y_1)$

$动直线l过定点(0,t)交椭圆\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1于两点(x_1,y_1),(x_2,y_2)$
根据直线的两点式有：$\cfrac{y_1-t}{x_1}=\cfrac{y_2-t}{x_2}\Rightarrow x_1y_2-x_2y_1=t(x_1-x_2)$
构造$x_1y_2-x_2y_1的对偶式x_1y_2+x_2y_1=\cfrac{(x_1y_2)^2-(x_2y_1)^2}{x_1y_2-x_2y_1}$
$=\cfrac{x_1^2y_2^2-x_2^2y_1^2}{t(x_1-x_2)}=\cfrac{b^2(1-\cfrac{x_2^2}{a^2})^2x_1^2-b^2(1-\cfrac{x_1^2}{a^2})x_2^2}{t(x_1-x_2)}=\cfrac{b^2(x_1^2-x_2^2)}{t(x_1-x_2)}=\cfrac{b^2}{t}(x_1+x_2)$

$例1、已知双曲线\Gamma ：\cfrac{x^2}{4}-y^2=1,AB为左右顶点，设过定点T(4,0)的直线与双曲线$
$交于CD两点(不与AB重合)，记直线AC,BD的斜率为k_1,k_2, 证明\frac{k_1}{k_2}为定值。-\cfrac{1}{3}$
听耳畔秋风知乎
$解：设l_{AB}:x=my+4,A(-2,0),B(2,0),C(x_1,y_1)D(x_2,y_2);$
$k_1=k_{AC}=\cfrac{y_1}{x_1+2},k_2=k_{BD}=\cfrac{y_2}{x_2-2}$
$\cfrac{y_1}{x_1-4}=\cfrac{y_2}{x_2-4}{\color{Green} \Rightarrow y_1(x_2-4)=y_2(x_1-4)\Rightarrow x_1y_2-x_2y_1=4(y_2-y_1)}$
它的对偶式有：
${\color{Green}x_1y_2+x_2y_1=\cfrac{(x_1y_2)^2-(x_2y_1)^2}{x_1y_2-x_2y_1} =\cfrac{x_1^2y_2^2-x_2^2y_1^2}{ 4(y_2-y_1)}}$
$=\cfrac{x_1^2y_2^2-x_2^2y_1^2}{ 4(y_2-y_1)}=\cfrac{4(y_1^2+1)y_2^2-4(y_2^2+1)y_2^2}{4(y_2-y_1)}=y_2+y_1$
$\cfrac{k_1}{k_2}=\cfrac{y_1}{x_1+2} \cdot\cfrac{x_2-2}{y_2} =\cfrac{x_2y_1-2y_1}{x_1y_2+2y_2}$
$\begin{cases} x_1y_2-x_2y_1=4y_2-4y_1\quad \\x_1y_2+x_2y_1=y_2+y_1 \qquad \end{cases}\Rightarrow$
$2y_2x_1=5y_2-3y_1;2y_1x_2=-3y_2+5y_1\Rightarrow \cfrac{k_1}{k_2}=\cfrac{2x_2y_1-4y_1}{2x_1y_2+4y_2}=\cfrac{-3y_2+5y_1-4y_1}{5y_2-3y_1+4y1}=-\cfrac{1}{3}$

$例2、设AB为椭圆\cfrac{x^2}{16}+\cfrac{y^2}{6} =1的长轴，该椭圆的动弦PQ过C(2,0),但不过原点，$翊空知乎
$直线AP与QB相交于M,PB与AQ相交于点N。求直线MN的方程。x=8$
根据极点极线知识可知，$l_{MN}为C(2,0)关于椭圆的极线段，x=8$
$解：设P(x_1,y_1),Q(x_2,y_2),A(-4,0),B(4,0)$
$l_{PQ}:\cfrac{y_1}{x_1-2} =\cfrac{y_2}{x_2-2} \Rightarrow x_1y_2-x_2y_1=2(y_2-y_1)$
容易得到它的对偶式：${\color{Red} x_1y_2+x_2y_1=8(y_2+y_1)} $
$\begin{cases} l_{AP}:x=\cfrac{x_1+4}{y_1} \cdot y-4 \quad①\\l_{BQ}:x=\cfrac{x_2-4}{y_2} \cdot y+4 \quad②\end{cases}$
$消y解出x_M$
$(x+4)\cdot \cfrac{y_1}{x_1+4}=(x-4)\cdot \cfrac{y_2}{x_2-4}\Rightarrow ( \cfrac{y_1}{x_1+4}-\cfrac{y_2}{x_2-4})\cdot x=-4( \cfrac{y_1}{x_1+4}+\cfrac{y_2}{x_2-4})\Rightarrow$
$x_M=\cfrac{-4(\cfrac{y_1}{x_1+4}+ \cfrac{y_2}{x_2-4})}{ \cfrac{y_1}{x_1+4}-\cfrac{y_2}{x_2-4}}=\cfrac{-4[y_1(x_2-4)+ y_2(x_1+4)]}{ y_1(x_2-4)-y_2(x_1+4)} =\cfrac{-4(x_1y_2+x_2y_1+4y_2-4y_1)}{x_2y_1-x_1y_2-4(y_1+y_2)}$
将对偶式代入上式，得$x_M=\cfrac{-4(8y_1+8y_2+4y_2-4y_1)}{2(y_1-y_2)-4y_1-4y_2}=8$

$例3、已知椭圆C：\cfrac{x^2}{a^2} +\cfrac{y^2}{b^2} =1(a\gt b\gt b\gt 0)过点P(2,\sqrt{2} )，$择梦周知乎
$离心率e为\cfrac{\sqrt{2} }{2} ，$
$1、求椭圆方程；$
$2、C的上下顶点为A,B,过点(0,4)斜率为k的直线与椭圆交于MN两点，证明直线BM与AN的$
$交点G在定直线，并求出该定直线方程。y=1$
https://one.free.nf/index.php/archives/43/
例4.2020年新课标I
$已知A,B分别为椭圆E:\cfrac{x^2}{a^2}+y^2=1(a>1)$左右两个顶点，G为E的上顶点，$\vec{AG} \cdot\vec{GB}=8.P为直线x=6上的动点，PA与E的另一交点为C,PB与E的另一交点为D.$

(1)求E的方程；$\cfrac{x^2}{9}+y^2=1 $
(2)证明：直线CD过定点。
$这题目是已知\frac{k_2}{k_2} =3求动直线过定点，根据极点极线的知识容易得到极点坐标为(\cfrac{3}{2} ,0)$

$解：设C(x_1,y_1)D(x_2,y_2),A(-3,0)B(3,0),P(6,t)$
$k_1=k_{AC}=\cfrac{y_1}{x_1+3}, k_2=k_{BD}=\cfrac{y_2}{x_2-3}$
$显然k_2=3k_1$
$预备知识\begin{cases} \cfrac{x_1^2}{a^2}+ \cfrac{y_1^2}{b^2}=1\\ \cfrac{x_2^2}{a^2}+ \cfrac{y_2^2}{b^2}=1\end{cases}$
两式相减，得$\cfrac{y_1-y_2}{x_1-x_2} =(e^2-1)\cfrac{x_1+x_2}{y_1+y_2} =-\cfrac{b^2}{a^2} \cfrac{x_1+x_2}{y_1+y_2}$
说明：这式子也是椭圆的第三定义的应用。是两点在椭圆上的斜率变换，加上两点的斜率公式，称作斜率双用。
$AC,BD各利用上面的变换，得 \begin{cases} k_1=k_{AC}=\cfrac{y_1}{x_1+3}=(e^2-1)\cfrac{x_1-3}{y_1}, ①\\ k_2=k_{BD}=\cfrac{y_2}{x_2-3}=(e^2-1)\cfrac{x_2+3}{y_2},②\end{cases}$
$①中左边的三倍＝②的左边；①中右边的三倍＝②的右边；$
$ \Rightarrow \begin{cases} 3\cfrac{y_1}{x_1+3}=\cfrac{y_2}{x_2-3} ,{\color{Red} ③} \\ \quad \\3\cfrac{x_1-3}{y_1}=\cfrac{x_2+3}{y_2},{\color{Red} ④} \end{cases}$
下面将有两种不同的方法得到答案。第一种是应用合比，第二种是斜率的对偶式应用。
$先来第一种：两式③④变形$
$\Rightarrow \begin{cases} \cfrac{y_1}{y_2}=\cfrac{(x_1+3)}{3(x_2-3)} , \\ \quad \\\cfrac{3(x_1-3)}{x_2+3}=\cfrac{y_1}{y_2},\end{cases}$
$\cfrac{y_1}{y_2}=\cfrac{4x_1-6}{4x_2-6}=\cfrac{x_1-\cfrac{3}{2} }{x_2-\cfrac{3}{2}}\Rightarrow {\color{Red} \cfrac{y_1}{x_1-\cfrac{3}{2} }=\cfrac{y_2}{x_2-\cfrac{3}{2}}}$
$故得，CD恒过(\cfrac{3}{2},0)$
法一毕！
法二前置知识：由直线的两点公式变形：
$\cfrac{y_1-{\color{Red} y} }{x_1-{\color{Red} x} } ＝\cfrac{y_1-y_2}{x_1-x_2}\Rightarrow 交叉相乘得\Rightarrow ( y_1-{\color{Red} y} )(x_1-x_2)=(x_1-{\color{Red} x} )(y_1-y_2)$
$\Rightarrow y_1(x_1-x_2)-{\color{Red} y} (x_1-x_2)=x_1(y_1-y_2)-{\color{Red} x} (y_1-y_2)$
$\Rightarrow y_1(x_1-x_2)-x_1(y_1-y_2)={\color{Red} y} (x_1-x_2)-{\color{Red} x} (y_1-y_2)$
$\Rightarrow x_1y_2-x_2y_1={\color{Red} y} (x_1-x_2)-{\color{Red} x} (y_1-y_2)$
$\Rightarrow x_1y_2-x_2y_1={\color{Red} y} (x_1-x_2)+{\color{Red} x} (y_2-y_1)\quad{\color{Red} \bullet \circ }$
法二正题：
$\begin{cases} 3\cfrac{y_1}{x_1+3}=\cfrac{y_2}{x_2-3} ,{\color{Red} ③} \\ \quad \\3\cfrac{x_1-3}{y_1}=\cfrac{x_2+3}{y_2},{\color{Red} ④} \end{cases}两式交叉相乘$
$\begin{cases} 3y_1(x_2-3)=y_2(x_1+3),{\color{Red} ③} \\ \quad \\3y_2(x_1-3)=y_1(x_2+3),{\color{Red} ④} 　\end{cases}观察这两式子特点。$

$\Rightarrow \begin{cases} 3y_1x_2-9y_1=y_2x_1+3y_2, \\ \quad \\3y_2x_1-9y_2=y_1x_2+3y_1,　\end{cases}$接着对两式作差！不是和！！
$\Rightarrow 3y_1x_2-3y_2x_1-9y_1+9y_2=y_2x_1+3y_2-y_1x_2-3y_1, $
$\Rightarrow 4y_1x_2-4y_2x_1=6y_1-6y_2\Rightarrow y_1x_2-y_2x_1=\cfrac{3}{2} (y_1-y_2)$
$对比\quad x_1y_2-x_2y_1={\color{Red} y} (x_1-x_2)+{\color{Red} x} (y_2-y_1)\quad{\color{Red} \bullet \circ }可知 ，CD恒过(\cfrac{3}{2},0)$
$例5、已知椭圆C：\cfrac{x^2}{a2}+\cfrac{y^2}{b^2} =1(a\gt b\gt0 )的离心率为\cfrac{\sqrt{2} }{2},且过点A(2,1)$