一、蒙日圆与双切线问题：
圆锥曲线的正交切线交点轨迹
在椭圆中，任意两条互相垂直的切相关的交点都在同一个圆上，它的圆心是椭圆中心，半径等于长、短半轴平方和的算术平均根，这个圆就是蒙日圆，也叫准圆。
$(1)椭圆\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1的正交切线交点轨迹为x^2+y^2=a^2+b^2;$
$(2)双曲线\cfrac{x^2}{a^2}-\cfrac{y^2}{b^2}=1的正交切线交点轨迹为x^2+y^2=a^2-b^2;$
$(3)抛物线y^2=2px的正交切线交点轨迹为准线方程x=-\cfrac{p}{2};$
以上反之也成立。

$设P(x_0,y_0),A(x_1,y_1),B(x_2,y_2)$
$设PA:y=k_1(x-x_0)+y_0;\qquad PB:PA:y=k_2(x-x_0)+y_0;\quad k_1k_2=-1$
$\begin{cases} y=k_1(x-x_0)+y_0\\ \quad \\ \cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1\end{cases}\Rightarrow \begin{cases} y=k_1x+{\color{Red} (y_0-k_1x_0)}\\ \quad \\ \cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1\end{cases}$
$\Rightarrow b^2x^2+a^2y^2-a^2b^2=0\Rightarrow b^2x^2+a^2[k_1(x-x_0)+y_0]^2-a^2b^2=0$
$\Rightarrow b^2x^2+a^2[k_1x+{\color{Red} (y_0-k_1x_0)} ]^2-a^2b^2=0$
$\Rightarrow b^2x^2+a^2[k_1^2x^2+2k_1{\color{Red} (y_0-k_1x_0)} x+{\color{Red} (y_0-k_1x_0)} ^2]-a^2b^2=0$
$\Rightarrow(a^2k_1^2+b^2)x^2+2k_1a^2{\color{Red} (y_0-k_1x_0)} x+a^2{\color{Red} (y_0-k_1x_0)} ^2-a^2b^2=0$

$\because \Delta =4k_1^2a^4(y_0-kx_0)^2-4a^2(a^2k_1^2+b^2)[(y_0-k_1x_0)^2-b^2]=0$
$\Rightarrow k_1^2a^2(y_0-kx_0)^2=(a^2k_1^2+b^2)[(y_0-k_1x_0)^2-b^2]$
$\Rightarrow k_1^2a^2(y_0-kx_0)^2=a^2k_1^2(y_0-k_1x_0)^2-a^2b^2k_1^2+b^2(y_0-k_1x_0)^2-b^4$
$\Rightarrow a^2b^2k_1^2+b^4=b^2(y_0-k_1x_0)^2$
$\Rightarrow a^2k_1^2+b^2=(y_0-k_1x_0)^2$
同理，联立 $\begin{cases} y=k_2(x-x_0)+y_0\\ \quad \\ \cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1\end{cases}\Rightarrow \bigtriangleup =0$解得$\Rightarrow a^2k_2^2+b^2=(y_0-k_2x_0)^2$
$显然k_1,k_2是关于k的一元二次方程a^2k^2+b^2=(y_0-kx_0)^2的两个根$
$a^2k^2+b^2=(y_0-kx_0)^2\Rightarrow (a^2-x_0^2)k^2+2x_0y_0k+b^2-y_0^2$
$\Rightarrow {\color{Red} k_1k_2=\cfrac{b^2-y_0^2}{a^2-x_0^2} =-1}\Rightarrow {\color{Red} b^2-y_0^2+a^2-x_0^2 =0 \Rightarrow x_0^2+y_0^2=a^2+b^2} $

$题目1、c在平面直角坐标系XOY中，已知圆O:x^2+y^2=25，圆C:x^2+(y-1)^2=r^2(0\lt r \lt 3)，点P(-3,4)，$
$M,N为圆O上异于点P的两点，若直线PM,PN与圆C都相切。$
$求证：当r变化时，直线MN的斜率为定值。$
$设M(x_1,y_1),N(x_2,y_2),P(-3,4)PM:y=k_1(x+3)+4,PN:y=k_2(x+3)+4$
$PM与x^2+(y-1)^2=r^2相切， 则有：$
$\cfrac{\left | 3k_1+3 \right | }{\sqrt{1+k_1^2} }=r{\color{Red} \quad直线与圆相切用圆心到直线的距离=r；直线与椭圆相切用\Delta=0} $
$\Rightarrow 9(k_1+1)^2=r^2(1+k_1^2)$
$同理，PN与x^2+(y-1)^2=r^2相切，则有：\Rightarrow 9(k_2+1)^2=r^2(1+k_2^2)$
$故k_1,k_2是关于k的一元二次方程\Rightarrow 9(k+1)^2=r^2(1+k^2)的两个根$
$得：(9-r^2)k^2+18k+9-r^2=0$
$k_1k_2=1\Rightarrow {\color{Red} \cfrac{y_1-4}{x_1+3} \cfrac{y_2-4}{x_2+3} =1\Rightarrow y_1y_2-4(y_1+y_2)+16=x_1x_2+3(x_1+x_2)+9}$
$设MN :y=kx+m与圆O联立，\begin{cases}y=kx+m \\\quad\\x^2+y^2=25 \end{cases}$
$\Rightarrow (1+k^2)x^2+2kmx+m^2-25=0,$
$x_1+x_2=\cfrac{-2km}{1+k^2};\qquad x_1x_2=\cfrac{m^2-25}{1+k^2}$
$y_1+y_2=kx_1+m+kx_2+m=\cfrac{-2k^2m}{1+k^2}+2m=\cfrac{2m}{1+k^2}$
$y_1y_2=(kx_1+m)(kx_2+m)=k^2x_1x_2+km(x_1+x_2)+m^2=\cfrac{k^2(m^2-25)}{1+k^2}+\cfrac{-2k^2m^2}{1+k^2}+m^2$
$=\cfrac{k^2m^2-25k^2-2k^2m^2+k^2m^2+m^2}{1+k^2}=\cfrac{m^2-25k^2}{1+k^2}$
$代入上式得：\cfrac{m^2-25k^2}{1+k^2} -4\cfrac{2m}{1+k^2}+16=3\cfrac{-2km}{1+k^2}+\cfrac{m^2-25}{1+k^2}+9$
$\Rightarrow m^2-25k^2-8m+7(1+k^2)=-6km+m^2-25\Rightarrow 9k^2-3km+4(m-4)=0$
$(3k-4)[3k-(m-4)]=0,解得k=\cfrac{4}{3},k=\cfrac{m-4}{3}$

$题目2、2018年浙江高考，已知点P是y轴左侧(不含y轴)一点，抛物线C:y^2=4x上存在不同的两点A,B$
$满足PA,PB的中点均在C上。$
$(1)设AB中点为M，证明：PM垂直于y轴。$
$(2)若点P为半椭圆x^2+\cfrac{y^2}{4}=1(x\le 0)上的动点，求\triangle PAB面积的取值范围。$
$分析：设P(x_0,y_0),A(x_1,y_1),B(x_2,y_2),证2y_0=y_1+y_2即可$
$这里设AB的横坐标为x_1,x_2令问题变得得很复杂，不如直接利用AB在抛物线上这个约束条件。$
$即x_1=\cfrac{y_1^2}{4} ,x_2=\cfrac{y_2^2}{4}{\color{Red} \qquad点在抛物线上，只设一个坐标} $
$解：设P(x_0,y_0),A(\cfrac{y_1^2}{4},y_1),B(\cfrac{y_2^2}{4},y_2)$
$则PA的中点坐标(\cfrac{y_1^2}{8}+\cfrac{x_0}{2},\cfrac{y_1+y_0}{2})代入y^2=4x$
$\Rightarrow (\cfrac{y_0+y_1}{2})^2=4\cdot (\cfrac{y_1^2}{8} +\cfrac{x_0}{2} )\Rightarrow$
$y_{\color{Red} 1} ^2-2y_0y_{\color{Red} 1} +8x_0-y_0^2=0$
$同理PB中点可得：y_{\color{Red} 2} ^2-2y_0y_{\color{Red} 2} +8x_0-y_0^2=0$
$由此可见y_1,y_2是关于　y的方程{\color{Red} y} ^2-2y_0{\color{Red} y} +8x_0-y_0^2=0的两根。$
$y_1+y_2=2y_0,得证$

练习：
1、 已知圆Ｏ：$x^2+y^2=1$，若直线$y=kx+2$上总存在点Ｐ，使得过点Ｐ作圆Ｏ的两条切线相互垂直，则实数$k$的取值范围是：

由蒙日圆概念可知，P点的轨迹为$x^2+y^2=2$，由于P点又在直线$y=kx+2$上，即点P既在圆上又在直线上，也就是说直线与圆恒有公共点，即直线与圆相切或相交。依据点到直线的距离，可知圆心O到直线的距离$d\le r\Rightarrow \cfrac{2}{\sqrt{1+k^2}}\le \sqrt{2}\Rightarrow 1+k^2\ge 2 \Rightarrow k\ge 1或k\le -1$

2、 给定椭圆Ｃ：$\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1(a\gt b\gt 0)$，称圆心在原点O半径为$\sqrt{a^2+b^2}$的圆是椭圆C的准圆，若椭圆C的一个焦点为$F(\sqrt{2},0)$,其短轴上的一个端点到Ｆ的距离为$\sqrt{3}$.
(1)求椭圆Ｃ的方程和其准圆方程。$\quad\cfrac{x^2}{3}+y^2=1,x^2+y^2=4$
(2)点Ｐ是椭圆Ｃ的准圆上的动点，过点Ｐ作椭圆的切线$l_1,l_2$交准圆于点MN;
$①当点Ｐ为准圆与y轨正半轴的交点时，求直线l_1,l_2的方程并证明l_1\bot l_2；$
$②求证线段MN的长为定值。$

$解：①当直线l_1,l_2中有一条斜率不存在时，不妨设l_1$不存在斜率，$l_1:x=\pm \sqrt{3},当l_1：x=\sqrt{3}时，l_1$与准圆交于点$(\sqrt{3},1)(\sqrt{3},-1)$，此时$l_2$为$y=1(或y=-1)$，显然$l_1与l_24垂直，同理可证l_1:x=-\sqrt{3},直线l_1,l_2$垂直。
$②当直线l_1,l_2中有一条斜率存在时，设P(x_0,y_0),其中x_0^2+y_0^2=4$
$设经过点P(x_0,y_0)与椭圆相切的直线为y=k(x-x_0)+y_0$
$\begin{cases}x^2+3y^2-3=0\\y=k(x-x_0)+y_0\end{cases}\Rightarrow x^2+3(kx+y_0-kx_0)^2-3=0$
$\Rightarrow (1+3k^2)x^2+6k(y_0-kx_0)x+3(y_0-kx_0)^2-3=0$
$\Rightarrow \Delta =[6k(y_0-kx_0)]^2-4(1+3k^2)\cdot [3(y_0-kx_0)^2-3]=0$
$\Rightarrow 3k^2(y_0-kx_0)^2=(1+3k^2)\cdot [(y_0-kx_0)^2-1]\Rightarrow $
$3k^2(y_0-kx_0)^2=(y_0-kx_0)^2-1+3k^2(y_0-kx_0)^2-3k^2$
$\Rightarrow (y_0-kx_0)^2-1-3k^2=0整理为{\color{Red} 以k为主元}的代数式$
$(3-x_0^2)k^2+2x_0y_0k+1-y_0^2=0$
$设直线l_1，l_2的斜率k_1,k_2，因为l_1,l_2$与椭圆相切，所以$k_1,k_2$是满足方程$(3-x_0^2)k^2+2x_0y_0k+1-y_0^2=0$
$\Rightarrow k_1k_2=\cfrac{1-y_0^2}{3-x_0^2}=\cfrac{1+x_0^2-4}{3-x_0^2}=-1\Rightarrow l_1\perp l_2$

3、 2020年杭州学军中学高三月考已知抛物线$C_1:y^2=2px(p\gt 0).圆C_2:(x-1)^2+y^2=r^2(r\gt 0)$,抛物线$C_1上的点到其准线的距离的最小值为\cfrac{1}{4}.$
(1)求抛物线$C_1$的方程和其准线方程。
(2)点$P(2,y_0)是抛物线C_1$在第一象限内一点，过点Ｐ作圆$C_2$的两条切线分别交抛物线$C_1$于$A,B(A,B异于点P),问是否存在圆C_2使AB$恰为其切线，若存在，求出r值；若不存在，说明理由。
$解：①由抛物线C_1上的点到其准线的距离的最小值为\cfrac{1}{4}.\Rightarrow \cfrac{p}{2}=\cfrac{1}{4}\Rightarrow p=\cfrac{1}{2},y^2=x,准线x=-\cfrac{1}{4}$
$②由①可得P(2,\sqrt{2}),假设存在C_2使得AB恰为其切线，设A(y_1^2,y_1)B(y_2^2,y_2),则经过PA的直线方程为：$
$y-\sqrt{2}=\cfrac{y_1-\sqrt{2}}{y_1^2-2}(x-2),即x-(y_1+\sqrt{2})y+\sqrt{2}y_1=0$
${\color{Red} 抛物线可以用两点式写出直线方程，区别于椭圆和双曲线要设方程}$
$由C_2(1,0)到PA的距离为r,得\cfrac{|1+\sqrt{2}y_1|}{\sqrt{1+(y_1+\sqrt{2})^2}}=r\Rightarrow r^2[1+(y_1+\sqrt{2})^2]=(1+\sqrt{2}y_1)^2$
$化简，得(2-r^2)y_1^2+2\sqrt{2}(1-r^2)y_1+1-3r^2=0$
$同理，得(2-r^2)y_2^2+2\sqrt{2}(1-r^2)y_2+1-3r^2=0$
$所以，y_1,y_2是方程(2-r^2)y^2+2\sqrt{2}(1-r^2)y+1-3r^2=0的两个不相等实根$
$故有y_1+y_2=-\cfrac{2\sqrt{2}(1-r^2)}{2-r^2}\qquad y_1y_2=\cfrac{1-3r^2}{2-r^2}$
$故l_{AB}:y-y_1=\cfrac{y_1-y_2}{y_1^2-y_2^2}(x-y_1^2)\Rightarrow (y-y_1)(y_1+y_2)=x-y_1^2\Rightarrow x-(y_1+y_2)y+y_1y_2=0$
$由C_2(1,0)到l_{AB}的距离为r,得r=\cfrac{|1+y_1y_2|}{\sqrt{1+(y_1+y_2)^2}}所以(1+\cfrac{1-3r^2}{2-r^2})^2=r^2+r^2[-\cfrac{2\sqrt{2}(1-r^2)^2}{2-r^2}]$
$(3-4r^2)^2=r^2(2-r^2)^2+8r^2(1-r^2)^2,化简，得r^6-4r^4+4r^2-1=0,即(r^2-1)(r^4-4r^2+1)=0$
$经分析知0\lt r\lt 1,因此r=\cfrac{\sqrt{5}-1}{2}$

4、2019年天津二模.椭圆Ｃ：$\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1(a\gt b\gt 0)$的离心率为$\cfrac{\sqrt{3}}{2}$,直线y=1与椭圆Ｃ两交点距离为8.
①求椭圆C的方程。
②设$R(x_0,y_0)$是椭圆C上支点，由原点O向圆$(x-x_0)^2+(y-y_0)^2=4$引两条切线，分别交椭圆C于点P,Q，若OP,OQ的斜率存在，并分别记为$k_1,k_2$，求证:$k_1k_2$为定值。
③在(2)条件下，试问$\left | OP \right | ^2+\left | OQ\right | ^2$是否为定值？若是，求之，若不是，请说明理由。

解:$①\cfrac{x^2}{20}+\cfrac{y^2}{5}=1$
$②设OP、OQ的直线方程分别为y=k_1x,y_2=k_2x,由直线OP为圆R的切线，有\cfrac{|k_1x_0-y_0|}{\sqrt{1+k_1^2}}=2$
$\Rightarrow (x_0^2-4)k_1^2-2x_0y_0k_1+(y_0^2-4)=0$
$同理可得 (x_0^2-4)k_2^2-2x_0y_0k_2+(y_0^2-4)=0$
$\therefore k_1,k_2是关于k的方程 (x_0^2-4)k^2-2x_0y_0k+(y_0^2-4)=0的两个不相等实数根$
$x^2-4\ne 0,\Delta \gt 0,则k_1k_2=\cfrac{y_0^2-4}{x_0^2-4}由R(x_0,y_0)在椭圆上，即y_0^2=5-\cfrac{1}{4}x_0^2$
$\therefore k_1k_2=\cfrac{y_0^2-4}{x_0^2-4}=\cfrac{1-\cfrac{1}{4}x_0^2}{x_0^2-4}=-\cfrac{1}{4}$
$③设P(x_1,y_1),Q(x_2,y_2)\begin{cases}y=k_1x\\ \cfrac{x^2}{25}+\cfrac{y^2}{5}\end{cases}\Rightarrow \begin{cases}x_1^2=\cfrac{20}{1+4k_1^2}\\y_1^2=\cfrac{20k_1^2}{1+4k_1^2}\end{cases}\Rightarrow x_1^2+y_1^2=\cfrac{20(1+k_1^2)}{1+4k_1^2}$
$同理，得x_2^2+y_2^2=\cfrac{20(1+k_2^2)}{1+4k_2^2},由k_1k_2=-\cfrac{1}{4}$
$\left | OP \right | ^2+\left | OQ\right | ^2=x_1^2+y_1^2+x_2^2+y_2^2=\cfrac{20(1+k_1^2)}{1+4k_1^2}+\cfrac{20(1+k_2^2)}{1+4k_2^2}=\cfrac{20(1+k_1^2)}{1+4k_1^2}+\cfrac{20(1+\cfrac{1}{16k_1^2})}{1+\cfrac{1}{4k_2^2}}$
$=\cfrac{20(1+k_1^2)}{1+4k_1^2}+\cfrac{20(4k_1^2+\cfrac{1}{4})}{4k_1^2+1}=\cfrac{100k_1^2+25}{1+4k_1^2}=25$

练习：
1、设点$A(x_1,y_1)B(x_2,y_2)$是椭圆$\cfrac{x^2}{4}+y^2=1$上两点，若过点A,B且斜率分别为$-\cfrac{x_1}{4y_1},-\cfrac{x_2}{4y_2}$的两条直线交于点P，且直线OA,OB的斜率之积为$-\cfrac{1}{4},E(\sqrt{6},0),\left | PE\right |的最小值为(\qquad)$

2、已知抛物线$C_1:x^2=y,圆C_2:x^2+(y-4)^2=1$圆心为点M。
(1)求点$M$到抛物线$C_1$准线的距离。
(2)已知点$P是C_1$上一点(异于原点)，过点$P$作圆$C_2$的两条切线，交抛物线$C_1$于$A,B$两点，若过$M,P$的直线$l$垂直于A,B，求$l$的方程。

3、
4、
5、
6、

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

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

$例1、已知椭圆C：\cfrac{x^2}{4} +\cfrac{y^2}{3}=1，求焦点的极线；$
$例2、\cfrac{x^2}{4} -y^2=1，求直线l:\frac{\sqrt{3} }{4} x-y=1的极点；$
$例3、已知椭圆C：\cfrac{x^2}{4} +\cfrac{y^2}{3}=1，左右端点A(-2,0),B(2,0), 过焦点F(1,0)的直线$
$l_0交椭圆C于PQ两点，连接AP,BQ交于点K，求K的轨迹方程.$
$解：画图找自极三角形，可知K的轨迹方程是F(1,0)关于C的极线\Rightarrow \frac{x}{4} =1\Rightarrow x=4,$
$即可用极点极线猜答案后证。$
解题过程：
$设l_0过焦点的直线方程为:x=my+1,点P(x_1,y_1),Q(x_2,y_2),K(x,y)({\color{Red} 反设直线，m为斜率的倒数} )$
$则有：\begin{cases} l_{AP}:x=\cfrac{x_1+2}{y_1} y-2\\ l_{BQ}:x=\cfrac{x_2-2}{y_1} y+2\end{cases}$
$消元去y，解x_K,\cfrac{y_1}{x_1+2} (x+2)=y=\cfrac{y_2}{x_2-2} (x-2),得$
$(\cfrac{y_2}{x_2-2} -\cfrac{y_1}{x_1+2}){\color{Red} x} =2(\cfrac{y_1}{x_1+2} +\cfrac{y_2}{x_2+2})\Rightarrow {\color{Red} x} =\cfrac{2(\cfrac{y_1}{x_1+2} +\cfrac{y_2}{x_2+2})}{\cfrac{y_2}{x_2-2} -\cfrac{y_1}{x_1+2}} $
$x=2\times \cfrac{{\color{Red} x_1y_2+x_2y_1} -2(y_1-y_2)}{{\color{Red} x_1y_2-x_2y_1} +2(y_1+y_2)}{\color{Peach} 红色部份为对偶式，我们还要可以应用圆锥曲线的对偶式，简化计算} $
${\color{Peach} 根据前面的猜测，我们只需要证明此式=4即可} $
$即证\cfrac{{\color{Red} x_1y_2+x_2y_1} -2(y_1-y_2)}{{\color{Red} x_1y_2-x_2y_1} +2(y_1+y_2)}=2$
$即证{\color{Red} x_1y_2+x_2y_1} -2(y_1-y_2)=2({\color{Red} x_1y_2-x_2y_1} +2(y_1+y_2))$
$即证\cfrac{{\color{Red} x_1y_2+x_2y_1} -2(y_1-y_2)}{{\color{Red} x_1y_2-x_2y_1} +2(y_1+y_2)}=2$
$即证:x_1y_2-3x_2y_1+6y_1+2y_2=0,消元去x,$
$即证:2my_1y_2-3(y_1+y_2)=0\quad {\color{Red} (\ast )}$
$应用韦达定理，联立l_0和C，\begin{cases} 3x^2+4y^2-12=0\\ x=my+1\end{cases}\Rightarrow$
$3(my+1)^2+4y^2-12=0\Rightarrow (3m^2+4)y^2+6my-9=0$
$\begin{cases} y_1+y_2=\cfrac{-6m}{3m^2+4} \quad ①\\y_1y_2=\cfrac{-9}{3m^2+4} \qquad ②\end{cases}$
$代入上式得，3\cdot \cfrac{-6m}{3m^2+4} -2m\cdot\cfrac{-9}{3m^2+4} =0$
$由此可见，Ｋ的轨迹为x=4是成立的。$

$例4、已知椭圆C：\cfrac{x^2}{4} +\cfrac{y^2}{3}=1，左右端点A(-2,0),B(2,0), 设点Ｋ的直线l:x=4上运动，$
$连接AK,BK, 分别交椭圆于P,Q,　问：直线PQ是否过定点，若是，请求出定值。$
解题思路：
从初始变量点K出发，设置变量，接着写出直线KA,KB，联立椭圆方程，求出点PQ坐标。
再用PQ坐标写出直线方程，观察一下哪个坐标可以代入成为恒等式！
最后一步最麻烦，直线方程代入是有计算量的关键式子，看出恒等式更是无从下手。那么利用极点极线先猜后证可以提供帮助！
$设K(4,t)，则有\begin{cases} l_{AP}:x=\cfrac{6}{t}y-2\\l_{BQ}:x=\cfrac{2}{t}y+2 \end{cases}先联立C与l_{AP}，求出Ｐ的坐标$
$\begin{cases} l_{AP}:x=\cfrac{6}{t}y-2\\3x^2+4y^2-12=0 \end{cases}\Rightarrow $
$3(\cfrac{6}{t}y-2)^2+4y^2-12=0 \Rightarrow 3(\cfrac{36}{t^2}y^2-\cfrac{24}{t}y+4)+4y^2-12=0$
$\Rightarrow (\cfrac{3\times 4\times 9}{t^2}+4 )y^2+\cfrac{3\times 4\times (-6)}{t}y =0\Rightarrow$
$y[(\cfrac{27}{t^2} +1)y-\cfrac{18}{t} ]=0 解得y=0,或{\color{Red} y_P=\cfrac{18t}{t^2+27} }代回直线l_{AP},得{\color{Red} x_P=\cfrac{-2t^2+54}{t^2+27} } $
$同理，联立l_{BQ}与C，有3(\cfrac{2}{t}y+2 )+4y^2-12=0\Rightarrow 3(\cfrac{24}{t^2}y+\cfrac{8}{t}y+4)+4y^2-12=0$
$\Rightarrow (\cfrac{3}{t^2} +1)y^2+\cfrac{6}{t}y=0 \Rightarrow {\color{Red} y_Q=-\cfrac{6t}{t^2+3} }$
$代入直线方程得，{\color{Red} x_Q=-\cfrac{2t^2-6}{t^2+3} }$
${\color{Green} 就这样，我们利用参数t，利用将PQ的坐标表示出来了} $
${\color{Green} 再设P(x_1,y_1)Q(x_2,y_2)写出直线PQ方程} $
$x=\cfrac{x_1-x_2}{y_1-y_2} (y-1)+x_1=\cfrac{x_1-x_2}{y_1-y_2} y-\cfrac{x_1-x_2}{y_1-y_2}$
$=\cfrac{x_1-x_2}{y_1-y_2} y-\cfrac{{\color{Red} x_2y_1-x_1y_2} }{y_1-y_2}$
$我们已经猜出定点坐标是F(1,0),所以我们可判断\cfrac{{\color{Red} x_2y_1-x_1y_2} }{y_1-y_2}=1；下一步，我们将PQ坐标代入证明它＝１即可$
$x_P=\cfrac{-2t^2+54}{t^2+27}，y_P=\cfrac{18t}{t^2+27}$
$x_Q=-\cfrac{2t^2-6}{t^2+3}，y_Q=-\cfrac{6t}{t^2+3} ，$
$\cfrac{{\color{Red} x_2y_1-x_1y_2} }{y_1-y_2}＝\cfrac{-\cfrac{2t^2-6}{t^2+3}\cdot\cfrac{18t}{t^2+27} -\cfrac{-2t^2+54}{t^2+27}\cdot -\cfrac{6t}{t^2+3} } {\cfrac{18t}{t^2+27}+\cfrac{6t}{t^2+3} }$
$=\cfrac{(-2t^2-6)\cdot18t -(-2t^2+54)\cdot (-6t)} {18t(t^2+3)+6t(t^2+27)}=$
$=\cfrac{3(2t^2-6)+54-2t^2}{3(t^2+3)+t^2+27} = \cfrac{4t^2+54-18}{4T^2+27+9} =1$
$由此可知直线PQ过点(1,0)$

不过这个解法太复杂了，完全依靠大力出奇迹。复杂点在于，用t表示点PQ的求解，和直线PQ直线的表示。既然难表示，那么就祭出设而不求大法！
法二：
$设直线l_{PQ}:x=my+n,点P(x_1,y_1)Q(x_2,y_2)写出两条直线$
$\begin{cases} l_{AP}: x=\cfrac{x_1+2}{y_1}y-2=m_1y-2\quad m_1=\cfrac{x_1+2}{y_1}\\ l_{BQ}:x=\cfrac{x_2-2}{y_1}y+2=m_2y+2\quad m_2=\cfrac{x_2-2}{y_2}\end{cases}$
$\frac{m_1}{m_2} =3=\cfrac{\cfrac{x_1+2}{y_1}}{\cfrac{x_2-2}{y_2}} \Rightarrow \cfrac{x_1+2}{y_1}=3\cfrac{x_2-2}{y_2}$
$\Rightarrow y_2(x_1+2)=3y_1(x_2-2)\Rightarrow x_1y_2-3x_2y_1+6y_1+2y_2=0\quad消元去x$
${\color{Red} 注：这里也可以不消元，而用斜率双用+直线两点式的变形求之}$
$2my_1y_2+(3n-6)y_1-(n+2)y_2=0\quad {\color{Red} (\ast )}$
$此式若n=1,便是例3的2my_1y_2-3(y_1+y_2)=0\quad {\color{Red} (\ast )}$
$马上上韦达定理，联立l_{PQ}和C$
$\begin{cases} x=my+n\qquad \qquad \\ 3x^2+4y^2-12=0\end{cases}\Rightarrow (3m^2+4)y^2+6mny+3n^2-12=0$
$\begin{cases} y_1+y_2=\cfrac{-6mn}{3m^2+4} \quad ①\\y_1y_2=\cfrac{3n^2-12}{3m^2+4}\quad②\end{cases}$
$②{\div} ①,\cfrac{y_1y_2}{y_1+y_2} =\cfrac{3n^2-12}{-6mn} =\cfrac{n^2-4}{-2mn} $
${\color{Red} \Rightarrow} y_1y_2={\color{Red} \cfrac{n^2-4}{-2mn}} \cdot (y_1+y_2){\color{Red} \Rightarrow} 2my_1y_2={\color{Red} \cfrac{n^2-4}{-n}} \cdot (y_1+y_2)$
${\color{Red} \cfrac{n^2-4}{-n}} \cdot (y_1+y_2)+(3n-6)y_1-(n+2)y_2=0\Rightarrow $
$(n^2-3n+2)y_1=(n^2+n-2)y_2\Rightarrow (n-2)(n-1)y_1=(n+2)(n-1)y_2$
$因为y_1\ne y_2,要上式恒成立，必要n=1,故直线过定点(1,0)$

10-19不消元，直接用斜率双用+直线两点式的变形
${\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} $
$\begin{cases}k_{AP}=\cfrac{y_1}{x_1+2}=-\cfrac{3}{4}\cdot\cfrac{x_1-2}{y_1}\\ \quad \\k_{BP}=\cfrac{y_2}{x_2-2}=-\cfrac{3}{4}\cdot\cfrac{x_2+2}{y_2}\end{cases}$
${\color{Red}\because \quad } 3k_{AP}=k_{BP}\Rightarrow \begin{cases}3 \cdot\cfrac{y_1}{x_1+2}=\cfrac{y_2}{x_2-2}\\ \quad \\3\cdot\cfrac{x_1-2}{y_1}=\cfrac{x_2+2}{y_2}\end{cases}\Rightarrow \begin{cases} 3y_1(x_2-2)=y_2(x_1+2) \\ \quad \\3y_2(x_1-2)=y_1(x_2+2)\end{cases}$
$\Rightarrow \begin{cases} 3x_2y_1-6y_1=x_1y_2+2y_2 \\ \quad \\3x_1y_2-6y_2=x_2y_1+2y_1　\end{cases}两式相减\Rightarrow　3x_1y_2-3x_2y_1+6(y_1-y_2)=x_2y_1-x_1y_2+2(y_1-y_2)$
$\Rightarrow4x_1y_2-4x_2y_1=4(y_2-y_1)\Rightarrow x_1y_2-x_2y_1=y_2-y_1$
比较两点式的变形，可知，直线过定点$(1,0)$
$\cfrac{y_1-{\color{Red} y} }{x_1-{\color{Red} x} } ＝\cfrac{y_1-y_2}{x_1-x_2}\Rightarrow 交叉相乘得\Rightarrow ( y_1-{\color{Red} y} )(x_1-x_2)=(x_1-{\color{Red} x} )(y_1-y_2)$
$\Rightarrow y_1(x_1-x_2)-{\color{Red} y} (x_1-x_2)=x_1(y_1-y_2)-{\color{Red} x} (y_1-y_2)$
$\Rightarrow x_1y_2-x_2y_1={\color{Red} y} (x_1-x_2)+{\color{Red} x} (y_2-y_1)\quad{\color{Red} \bullet \circ }$