答案:(1)椭圆  \Gamma  的方程为  \frac{x^{2}}{4}+\frac{y^{2}}{3}=1 ,(2)提示;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\\(2)证明：(i)若直线  A B  的斜率不存在,设直线  A B  的方程为  x=t ,由对称性可知  t= \pm 1  ,\\则  |A B|=\left|y_{1}-y_{2}\right|=3, S_{\triangle A O B}=\frac{3}{2}  ,平行四边形  O A C B  的面积为 3 .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
(ii)若直线  A B  的斜率存在,设直线  A B  的方程为  y=k x+m  ,联立  \left\{\begin{array}{l}y=k x+m \\ 3 x^{2}+4 y^{2}=12\end{array}\right.  ,~~
\\消元  y  整理得  \left(4 k^{2}+3\right) x^{2}+8 k m x+4\left(m^{2}-3\right)=0  ,则  \Delta=48\left(4 k^{2}+3-m^{2}\right)>0, ~~~~~\\
\boxed{x_{1}+x_{2}=-\frac{8 k m}{4 k^{2}+3} } ,  x_{1} x_{2}=\frac{4\left(m^{2}-3\right)}{4 k^{2}+3}, \boxed{y_{1}+y_{2}=k\left(x_{1}+x_{2}\right)+2 m=\frac{6 m}{4 k^{2}+3}  },~~~~~~
\\\because OACB为平行四边形，\boxed{\overrightarrow{OA} +\overrightarrow{OC} =\overrightarrow{OB} ,\therefore B在椭圆上},
代入椭圆  \Gamma  的方程, ~~~~~~~~~~~~~~\\

整理得 \boxed{ 4 k^{2}+3=4 m^{2}}  ,于是  S_{\triangle A O B}=\frac{1}{2}|m|\left|x_{1}-x_{2}\right|=
\frac{1}{2}|m| \cdot \frac{\sqrt{48 \cdot 3 m^{2}}}{4 m^{2}}=\frac{3}{2}  ,则平行\\四边形  O A C B  的面积为 3 .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

答案:(1)\frac{x^2}{4}+\frac{y^2}{3}=1(y\ne 0),(2)[12,8\sqrt{3});~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
提示:(1)|EB|=|ED|，所以|EA|+|EB|=4>|AB|,所以,E点轨迹为椭圆.~~~~~~~~~~~~\\
(2)因为直线  l  的斜率不为 0 ，设直线  l  的方程为  x=m y+1 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
因为  l \perp P Q  ，设直线  P Q  的方程为  y=-m(x-1).设M\left(x_{1}, y_{1}\right), ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\

由\begin{cases}x = my + 1\\3x^{2}+4y^{2}=12\end{cases}可得(3m^{2}+4)y^{2}+6my - 9 = 0.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
可得y_{1}+y_{2}=-\frac{6m}{3m^{2}+4},y_{1}y_{2}=-\frac{9}{3m^{2}+4}.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\\则\vert MN\vert=\sqrt{1 + m^{2}}\cdot\vert y_{1}-y_{2}\vert=\frac{12\left(m^{2}+1\right)}{3 m^{2}+4}  ,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\又圆心  A  到  P Q  距离  d=\frac{|2 m|}{\sqrt{m^{2}+1}}  ，所以  |P Q|=2 \sqrt{4^{2}-d^{2}}=   \frac{4 \sqrt{4+3 m^{2}}}{\sqrt{1+m^{2}}}, ~~~~~~~~~~~~~~~~~\\
S_{\text {四边形 } M P N Q}=\frac{1}{2}|M N| \cdot|P Q|=\frac{1}{2} \cdot   \frac{12\left(m^{2}+1\right)}{3 m^{2}+4} \cdot \frac{4 \sqrt{3 m^{2}+4}}{\sqrt{m^{2}+1}}=24 \sqrt{\frac{1}{3+\displaystyle\frac{1}{m^{2}+1}}}  .~~\\
当m = 0时,S取得最小值12,又\frac{1}{1 + m^{2}}>0,可得S<8\sqrt{3},所以S\in[12,8\sqrt{3})~~~~~~~~~~

答案:(1)  \frac{x^{2}}{3}-y^{2}=1  .
(2)证明:①若  l  与  x  轴不重合,设  l: x=t y+3  ,
由  \left\{\begin{array}{l}x=t y+3, \\ \displaystyle\frac{x^{2}}{3}-y^{2}=1,\end{array}\right.  \\得  \left(t^{2}-3\right) y^{2}+6 t y+6=0  ,
所以  \left\{\begin{array}{l}t^{2}-3 \neq 0, \\ \Delta=(6 t)^{2}-24\left(t^{2}-3\right)>0,\end{array}\right.    y_{1}+y_{2}=-\frac{6 t}{t^{2}-3}, \\y_{1} y_{2}=\frac{6}{t^{2}-3}  .
又直线  A Q  的方程为  y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-1}(x-1)  ,
由对称性可知，直线l过的定\\点在x 轴，令y=0,.\boxed{x_0=-  \frac{y_{1}\left(t y_{2}+2\right)}{y_{2}-y_{1}}=\frac{t y_{1} y_{2}+2 y_{1}}{y_{2}-y_{1}}+1=2}.直线  A Q  过定点  (2,0)  .~~~\\
②当  l  与  x  轴重合时,直线  A Q  过定点  (2,0)  .综上,直线  A Q  过定点  (2,0)  .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

(1) l_1\perp x轴,\Rightarrow A(2,2),B(2, - 2),S=\frac{1}{2}|AB|\vert x_C - x_D\vert=4\sqrt{5}
设l_2:y = k(x - 2),~~~~~~~~
\\联立y^{2}=2x\Rightarrow\vert x_C - x_D\vert=\frac{\sqrt{16k^{2}+4}}{k^{2}} = 2\sqrt{5}
\Rightarrow k^{2}=1或k^{2}=-\frac{1}{5}(舍)~~~~~~~~~~~~~~~~~~~~~~~ \\
 \therefore l_2:y=\pm(x - 2).~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
 
(2) 由题意AD,BC关于x轴对称,\therefore AC,BD交点在x轴,
设AC:y = kx + m,~~~~~~~~~~~~~~~~~~~\\
联立y^{2}=2x,\Rightarrow ky^{2}-2y + 2m = 0,
y_1 + y_2=\frac{2}{k},y_1y_2=\frac{2m}{k}.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\k_{AN}+k_{CN}=0,\Rightarrow\frac{y_1}{x_1 - 2}+\frac{y_2}{x_2 - 2}=0,由x=\frac{y^2}{2}
\Rightarrow y_1y_2 = 4=\frac{2m}{k},\therefore m = 2k  ~~~~~~\\\therefore AC:y=kx + 2k,\therefore AC过(-2,0)
\therefore AC,BD交点M(-2,0),~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\\because MA,MD关于x轴对称,
\therefore\triangle MAB的圆心Q在x轴上,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
设Q(t,0),A(x_1,y_1)
\therefore\frac{\vert MQ\vert}{\vert NQ\vert}=\frac{\vert MA\vert}{\vert NA\vert}\Rightarrow\frac{t + 2}{2-t}=\frac{\sqrt{(x_1 + 2)^{2}+y_1^{2}}}{\sqrt{(x_1 - 2)^{2}+y_1^{2}}}=\sqrt{\frac{x_1^{2}+6x_1 + 4}{x_1^{2}-2x_1 + 4}}
=\\\sqrt{1+\frac{8x_1}{x_1^{2}-2x_1 + 4}}=\sqrt{1+\frac{8}{x_1+\frac{4}{x_1}-2}}\in(1,\sqrt{5})
\Rightarrow t\in(0,3-\sqrt{5}).~~~~~~~~~~~~~~~~~~~~~~~~~~~

答案:(1)\frac{x^{2}}{8}+\frac{y^{2}}{2}=1;(2)不存在；~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\\提示:法一、设角平分线l与x轴交于点M(t,0),

\frac{|AF_{1}|}{|AF_{2}|}=\frac{|F_{1}M|}{|F_{2}M|}~~~★角平分线定理★~~ \\\Rightarrow \frac{(2 + \sqrt{6})^{2}+1}{(2 - \sqrt{6})^{2}+1}=\frac{(t + \sqrt{6})^{2}}{(t - \sqrt{6})^{2}} \Rightarrow t = \frac{3}{2}, \therefore M(\frac{3}{2},0)
\therefore k_{AM}=2, \therefore \text{直线}l: y = 2x - 3.\\
设直线l的垂线为l_{1}: y =-\frac{1}{2}x + m,
\left\{
\begin{array}{l}
y =-\displaystyle \frac{1}{2}x + m \\
x^{2}+4y^{2}=8
\end{array}
\right. \Rightarrow 2x^{2}-4mx + 4m^{2}-8 = 0, \\\Delta>0, \Rightarrow - 2\leq m<2,
x_{1}+x_{2}=2m = 2x_{0}, \therefore x_{0}=m, y_{0}=\frac{m}{2}, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\\text{将}(m,\frac{m}{2})\text{代入}y = 2x - 3,
\Rightarrow m = 2, \text{不符合}\Delta>0, \therefore \text{不存在}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\\法二、★光学性质★~~~过点A的切线方程为:\frac{2x}{8}+\frac{y}{2}=1,过点A的垂线:y=2x-3.

答案:(1) y^{2}=2x,(2) (ⅰ) 提示,(ⅱ) 有2条;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
提示:

(2)(ⅰ)设l:x = my + 2,\therefore k_{AB}=\frac{1}{m},~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
\begin{cases}x = my + 2\\y^{2}=2x\end{cases}\Rightarrow y^{2}-2my - 4 = 0,
y_{1}+y_{2}=2m,y_{1}y_{2}=-4,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
直线AO:y=\frac{y_{1}}{x_{1}}x\Rightarrow D(-2,\frac{-2y_{1}}{x_{1}}),
直线BE:y = x + x_{2}\Rightarrow E(-x_{2},0),~~~~~~~~~~~~~~~~~~~\\
k_{DE}=\frac{2y_{1}}{x_{1}(-2 - x_{2})},代入:x_{1}=\frac{y_{1}^{2}}{2},x_{2}=my_{2}+2,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
\Rightarrow k_{DE}=-\frac{4}{my_{1}y_{2}}=\frac{1}{m}=k_{AB}\Rightarrow DE//~l~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\\(ⅱ) S_{四边形ABDE}
 = S_{DBME}+S_{\triangle AEM}=(2 - x_{E})(|y_{B}|+\frac{1}{2}|y_{A}|)=(2+\frac{y_{2}^{2}}{2})(-y_{2}+\frac{1}{2}y_{1})\\
=-\frac{1}{2}y_{2}^{3}-3y_{2}-\frac{y_{2}}{y_{1}},(y_{2}<0),
S = 12\Rightarrow\frac{1}{2}y_{2}^{4}+3y_{2}^{2}+12y_{2}+4 = 0,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
令f(x)=\frac{1}{2}x^{4}+3x^{2}+12x + 4,x<0,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
f'(x)=2x^{3}+6x + 12,
f''(x)=6x^{2}+6>0\Rightarrow f'(x)单增,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\且f'(-2)=-16<0,f'(-1)>0
\because\exists x_{0}\in(-2,-1),使f'(x_{0}) = 0,f(x)极小值为f(x_{0}),\\
又\because f(-2)=0,f(-1)<0,f(0)>0 \therefore f(x)=0有2解.
\therefore l有2条.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

答案: (1) \frac{x^{2}}{4}+y^{2}=1,(2) \sqrt{5};~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
提示:
(2) 设l:y = \frac{1}{2}x + m,\begin{cases}y = \displaystyle \frac{1}{2}x + m\\\displaystyle \frac{x^{2}}{4}+y^{2}=1\end{cases}\Rightarrow x^{2}+2mx + 2m^{2}-2 = 0,~~~~~~~~~~~~~~~~~~~~~~\\
\Delta>0\Rightarrow\Delta = 8 - 4m^{2}>0,\Rightarrow-\sqrt{2}< m< \sqrt{2},
x_{1}+x_{2}=-2m,x_{1}x_{2}=2m^{2}-2.~~~\\
GH 的中点为圆心M(-m,\frac{m}{2}),
\vert GH\vert=\sqrt{1 + \frac{1}{4}}\cdot\sqrt{8 - 4m^{2}}=\sqrt{5}\cdot\sqrt{2 - m^{2}} = 2R,~~\\
\vert ON\vert_{max}=\vert OM\vert+R=\frac{\sqrt{5}}{2}\vert m\vert+\frac{\sqrt{5}}{2}\sqrt{2 - m^{2}}=\frac{\sqrt{5}}{2}(\sqrt{m^{2}}+\sqrt{2 - m^{2}})~~~~~~~~~~~~~~~~~\\
\leq\frac{\sqrt{5}}{2}\cdot2\sqrt{\frac{m^{2}+2 - m^{2}}{2}}=\sqrt{5},当且仅当m = \pm1时成立.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
★ab\leq(\frac{a + b}{2})^{2}\leq\frac{a^{2}+b^{2}}{2}\Rightarrow a + b\leq2\sqrt{\frac{a^{2}+b^{2}}{2}}★

 (1)\displaystyle \frac{x^{2}}{4}+y^{2}=1,(2)[\frac{8\sqrt{3}}{3},\infty );~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
(2)
{\color{Red} 法一、}设过点R(t,2)且与椭圆相切的直线为:~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
x - t = m(y - 2),即:\boxed{x = my + t - 2m}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
\begin{cases}x = my + t - 2m\\ \displaystyle \frac{x^{2}}{4}+y^{2}=1\end{cases}\Rightarrow (m^{2}+4)y^{2}+2m(t - 2m)y+(t - 2m)^{2}-4 = 0~~~~~~~~~~~~~~~~~~~~~~~\\
\Delta = 0，\Rightarrow 4m^{2}(t - 2m)^{2}-4(m^{2}+4)[(t - 2m)^{2}-4]=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
\Rightarrow m^{2}(t - 2m)^{2}-m^{2}[(t - 2m)^{2}-4]-4[(t - 2m)^{2}-4]=0]=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
★注意运算技巧，(t - 2m)^{2}作为整体★\\
\boxed{\Rightarrow3m^{2}-4tm + t^{2}-4 = 0}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
\Rightarrow m_1 + m_2=\frac{4t}{3}，m_1m_2=\frac{t^{2}-4}{3}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\

l_{RM}:x = m_1y + t - 2m_1,令y = 0得:M(t - 2m_1,0),
同理：N(t - 2m_2,0)~~~~~~~~~~~~~~~~\\
S_{MNR}=\frac{1}{2}\times2|MN|=2|m_1-m_2|=\sqrt{(m_1 + m_2)^{2}-4m_1m_2})
=2\sqrt{\frac{4t^{2}+48}{9}}\geqslant\frac{8\sqrt{3}}{3}.

{\color{Red}法二、}当l斜率存在时设切线为：y-2=k(x-t),当斜率不存在时，求面积。~~~~~~~~~~\\
★设直线的点斜式方程，与圆锥曲线相切。注意直线的形式。★

{\color{Red} 法三、}设点  M(m, 0), N(n, 0), R(t, 2)  ,注意到  R M, R N  斜率不为 0 ,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
设  R M: \boxed{x=\frac{t-m}{2} y+m}, R N: x=\frac{t-n}{2} y+n  ,联立  \left\{\begin{array}{l}\displaystyle \frac{x^{2}}{4}+y^{2}=1 \\ x=\displaystyle\frac{t-m}{2} y+m\end{array}\right.  ,~~~~~~~~~~~\\得  \left[16+(t-m)^{2}\right] y^{2}+4(t-m) m y+4\left(m^{2}-4\right)=0  ,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\  \Delta=[4(t-m) m]^{2}-4\left[16+(t-m)^{2}\right] \cdot 4\left(m^{2}-4\right)=0  ,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\于是  16(t-m)^{2} m^{2}-16\left[16+(t-m)^{2}\right]\left(m^{2}-4\right)=0  ,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\化简得  \boxed{3 m^{2}+2 t m-t^{2}-16=0}  ,

同理有  3 n^{2}+2 t n-t^{2}-16=0  ,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\故  m, n  是一元二次方程 \boxed{ 3 x^{2}+2 t x-t^{2}-16=0  }的两根,
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
则  m+n=-\frac{2 t}{3}, m n=-\frac{t^{2}+16}{3}  ,
所以  |M N|=|m-n|=\sqrt{(m+n)^{2}-4 m n}=~~~~~~~~\\\sqrt{\left(-\frac{2 t}{3}\right)^{2}-4 \times\left(-\frac{t^{2}+16}{3}\right)}=\frac{4 \sqrt{t^{2}+12}}{3}  ,
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
又  t^{2} \geqslant 0  ,所以  S_{\triangle R M N}=\frac{1}{2}|M N| \cdot 2=\frac{4 \sqrt{t^{2}+12}}{3} \geqslant \frac{4 \sqrt{0+12}}{3}=\frac{8 \sqrt{3}}{3}  .~~~~~~~~~~~~~~~~~~~~~~~~\\★设两点坐标，表达直线，注意直线的形式。★