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

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