$例1、已知椭圆C：\cfrac{x^2}{4} +\cfrac{y^2}{3}=1，求焦点的极线；$
$例2、\cfrac{x^2}{4} -y^2=1，求直线l:\frac{\sqrt{3} }{4} x-y=1的极点；$
$例3、已知椭圆C：\cfrac{x^2}{4} +\cfrac{y^2}{3}=1，左右端点A(-2,0),B(2,0), 过焦点F(1,0)的直线$
$l_0交椭圆C于PQ两点，连接AP,BQ交于点K，求K的轨迹方程.$
$解：画图找自极三角形，可知K的轨迹方程是F(1,0)关于C的极线\Rightarrow \frac{x}{4} =1\Rightarrow x=4,$
$即可用极点极线猜答案后证。$
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$(1,0)$
$\cfrac{y_1-{\color{Red} y} }{x_1-{\color{Red} x} } ＝\cfrac{y_1-y_2}{x_1-x_2}\Rightarrow 交叉相乘得\Rightarrow ( y_1-{\color{Red} y} )(x_1-x_2)=(x_1-{\color{Red} x} )(y_1-y_2)$
$\Rightarrow y_1(x_1-x_2)-{\color{Red} y} (x_1-x_2)=x_1(y_1-y_2)-{\color{Red} x} (y_1-y_2)$
$\Rightarrow x_1y_2-x_2y_1={\color{Red} y} (x_1-x_2)+{\color{Red} x} (y_2-y_1)\quad{\color{Red} \bullet \circ }$