圆锥曲线综合题-提高篇

题目

1.椭圆𝐶与双曲线2𝑥22𝑦2=1有相同的焦点,且过(1,32).                                               (1)求椭圆𝐶的方程.                                                                                                                   (2)如图所示,记椭圆的左、右顶点分别为𝐴,𝐵,当动点𝑀在直线𝑥=4上运动时,直线      𝐴𝑀,𝐵𝑀分别交椭圆于点𝑃,𝑄.                                                                                               𝑎.证明:点𝐵在以𝑃𝑄为直径的圆内;                                                                                     𝑏.求四边形𝐴𝑃𝐵𝑄面积的最大值.                                                                                             1.椭圆𝐶与双曲线2𝑥^2-2𝑦^2=1有相同的焦点,且过(1,\frac{3}{2}).~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (1)求椭圆𝐶的方程.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2)如图所示,记椭圆的左、右顶点分别为𝐴,𝐵,当动点𝑀在直线𝑥=4上运动时,直线~~~~~~\\𝐴𝑀,𝐵𝑀分别交椭圆于点𝑃,𝑄.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 𝑎.证明:点𝐵在以𝑃𝑄为直径的圆内;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 𝑏.求四边形𝐴𝑃𝐵𝑄面积的最大值.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
练习1:.已知椭圆Γ:x2a2+y2b2=1(a>b>0)的离心率为12,且经过点(1,32).               (1)求椭圆Γ的方程;                                                                                                                    (2)已知O为坐标原点,若平行四边形OACB的三个顶点A,B,C均在椭圆Γ,求证:平 行四边形OACB的面积为定值.                                                                                               练习1:.已知椭圆 \Gamma: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0) 的离心率为 \frac{1}{2},且经过点 \left(1,\frac{3}{2}\right) .~~~~~~~~~~~~~~~\\ (1)求椭圆 \Gamma 的方程;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2)已知 O 为坐标原点,若平行四边形 O A C B 的三个顶点 A, B, C 均在椭圆 \Gamma 上,求证:平~\\行四边形 O A C B 的面积为定值.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
参考图象

答案

答案:(1)椭圆Γ的方程为x24+y23=1,(2)提示;                                                                    (2)证明:(i)若直线AB的斜率不存在,设直线AB的方程为x=t,由对称性可知t=±1,AB=y1y2=3,SAOB=32,平行四边形OACB的面积为3.                                  (ii)若直线AB的斜率存在,设直线AB的方程为y=kx+m,联立{y=kx+m3x2+4y2=12,  消元y整理得(4k2+3)x2+8kmx+4(m23)=0,Δ=48(4k2+3m2)>0,     x1+x2=8km4k2+3,x1x2=4(m23)4k2+3,y1+y2=k(x1+x2)+2m=6m4k2+3,      OACB为平行四边形,OA+OC=OB,B在椭圆上,代入椭圆Γ的方程,              整理得4k2+3=4m2,于是SAOB=12mx1x2=12m483m24m2=32,则平行四边形OACB的面积为3.                                                                                                         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 .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
练习2.设圆x2+y2+2x15=0的圆心为A,直线l过点B(1,0)且与x轴不重合,l交     AC,D两点,BAC的平行线交AD于点E.                                                            (1)证明EA+EB为定值,并写出点E的轨迹方程;                                                      (2)设点E的轨迹为曲线C1,直线lC1M,N两点,过B且与l垂直的直线与圆A交    P,Q两点,求四边形MPNQ面积的取值范围.                                                                 练习2.设圆 x^{2}+y^{2}+2 x-15=0 的圆心为 A ,直线 l 过点 B(1,0) 且与 x 轴不重合, l 交~~~~~\\圆 A 于 C, D 两点,过 B 作 A C 的平行线交 A D 于点 E .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (1)证明 |E A|+|E B| 为定值,并写出点 E 的轨迹方程;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2)设点 E 的轨迹为曲线 C_{1} ,直线 l 交 C_{1} 于 M, N 两点,过 B 且与 l 垂直的直线与圆 A 交~~~~\\于 P, Q 两点,求四边形 M P N Q 面积的取值范围.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
参考图象

答案

答案:(1)x24+y23=1(y0),(2)[12,83);                                                                       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=my+1                                               因为lPQ,设直线PQ的方程为y=m(x1).M(x1,y1),                                     {x=my+13x2+4y2=12可得(3m2+4)y2+6my9=0.                                                      可得y1+y2=6m3m2+4,y1y2=93m2+4.                                                                    MN=1+m2y1y2=12(m2+1)3m2+4,                                                               又圆心APQ距离d=2mm2+1,所以PQ=242d2=44+3m21+m2,                 S四边形 MPNQ=12MNPQ=1212(m2+1)3m2+443m2+4m2+1=2413+1m2+1.  m=0,S取得最小值12,11+m2>0,可得S<83,所以S[12,83)          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})~~~~~~~~~~

2.已知椭圆𝐸的左、右焦点分别为𝐹1(c,0),𝐹2(c,0)(c>0),𝑀在椭圆𝐸上,              𝑀𝐹2𝐹1𝐹2,𝑀𝐹1𝐹2的周长为4+23,面积为12𝑐.                                                            (1)求椭圆𝐸的方程.                                                                                                                   (2)设椭圆𝐸的左、右顶点分别为𝐴,𝐵,过点(1,0)的直线𝑙与椭圆𝐸交于𝐶,𝐷两点(不同于左右顶点),记直线𝐴𝐶的斜率为𝑘1,直线𝐵𝐷的斜率为𝑘2,问是否存在实常数λ,使得𝑘1=λk2恒成立?若成立,求出λ的值,若不成立,说明理由.2.已知椭圆𝐸的左、右焦点分别为𝐹_1(-c,0),𝐹_2(c,0)(c>0),点𝑀在椭圆𝐸上,~~~~~~~~~~~~~~\\𝑀𝐹_2⊥𝐹_1𝐹_2, ∆𝑀𝐹_1𝐹_2的周长为4+2\sqrt{3},面积为\frac{1}{2}𝑐.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (1)求椭圆𝐸的方程.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2)设椭圆𝐸的左、右顶点分别为𝐴,𝐵,过点(1,0)的直线𝑙与椭圆𝐸交于𝐶,𝐷两点(不同于\\左右顶点),记直线𝐴𝐶的斜率为𝑘_1,直线𝐵𝐷的斜率为𝑘_2,问是否存在实常数λ,使得\\𝑘_1=λk_2恒成立?若成立,求出λ的值,若不成立,说明理由.\quad\quad\quad\quad\quad\quad\quad\quad\quad
练习1.已知双曲线C:x2a2y2b2=1(a>0,b>0),四点M1(4,23),M2(3,2),          M3(2,33)M4(2,33)中恰有三点在C.                                                           (1)C的方程;                                                                                                                        (2)过点(3,0)的直线lCP,Q两点,过点P作直线x=1的垂线,垂足为A.证明:直   线AQ过定点.                                                                                                                            练习1.已知双曲线 C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0) ,四点 M_{1}\left(4, \frac{\sqrt{2}}{3}\right), M_{2}(3, \sqrt{2}), ~~~~~~~~~~\\M_{3}\left(-2,-\frac{\sqrt{3}}{3}\right) , M_{4}\left(2, \frac{\sqrt{3}}{3}\right) 中恰有三点在 C 上.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (1)求 C 的方程;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2)过点 (3,0) 的直线 l 交 C 于 P, Q 两点,过点 P 作直线 x=1 的垂线,垂足为 A .证明:直~~~\\线 A Q 过定点.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案

答案:(1)x23y2=1.(2)证明:①若lx轴不重合,l:x=ty+3,{x=ty+3,x23y2=1,(t23)y2+6ty+6=0,所以{t230,Δ=(6t)224(t23)>0,y1+y2=6tt23,y1y2=6t23.又直线AQ的方程为yy1=y2y1x21(x1),由对称性可知,直线l过的定点在x轴,令y=0,.x0=y1(ty2+2)y2y1=ty1y2+2y1y2y1+1=2.直线AQ过定点(2,0).   ②当lx轴重合时,直线AQ过定点(2,0).综上,直线AQ过定点(2,0).                                   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) .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

3.已知椭圆W:x24m+y2m=1(m>0)的长轴长为4,左、右顶分别为A,B经过点          P(1,0)的动直线与椭圆W交于不同的两点C,D(不与点A,B重合)。                                (1)求椭圆W的方程及离心率;                                                                                               (2)若直线CB与直线AD相交于M,判断点M是否位于一条定直线上?若是,求出该    直线的方程,若不是说明理由。                                                                                            3.已知椭圆W:\frac{x^2}{4m}+\frac{y^2}{m}=1(m>0)的长轴长为4,左、右顶分别为A,B经过点~~~~~~~~~~\\P(1,0)的动直线与椭圆W交于不同的两点C,D(不与点A,B重合)。~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (1)求椭圆W的方程及离心率;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2)若直线CB与直线AD相交于M,判断点M是否位于一条定直线上?若是,求出该~~~~\\直线的方程,若不是说明理由。~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

4.已知椭圆C:x2a2+y2b2=1(a>b>0)的右焦点F与抛物线E:y2=2px(p>0)的焦    点相同,曲线C的离心率为12,M(2,y)E上的一点且MF=3.                                      (1)求曲线C和曲线E的方程。                                                                                                 (2)若直线l:y=kx+2交曲线CP,Q两点,ly轴于R.                                                 a.求三角形POQ面积的最大值(O为坐标原点)b.BP=λRQ,求实数λ的取值范围。4.已知椭圆C:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)的右焦点F与抛物线E:y^2=2px(p>0)的焦~~~~\\点相同,曲线C的离心率为\frac{1}{2},M(2,y)为E上的一点且|MF|=3.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ (1)求曲线C和曲线E的方程。~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ (2)若直线l:y=kx+2交曲线C与P,Q两点,l交y轴于R.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ a.求三角形POQ面积的最大值(O为坐标原点)。\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ b.\vec{BP}=λ\vec{RQ},求实数λ的取值范围。\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad
5.已知拋物线y2=2x,过点N(2,0)作两条直线l1,l2分别交拋物线于A,BC,D(其中    A,Cx轴上方).                                                                                                                      (1)l1垂直于x,且四边形ACBD的面积为45,求直线l2的方程;                                   (2)l1,l2倾斜角互补时,直线AC与直线BD交于点M,MAB的内切圆的圆心横坐 标的取值范围.                                                                                                                          5.已知拋物线 y^{2}=2 x ,过点 N(2,0) 作两条直线 l_{1}, l_{2} 分别交拋物线于 A, B 和 C, D (其中~~~~\\ A, C 在 x 轴上方).~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (1)当 l_{1} 垂直于 x 轴,且四边形 A C B D 的面积为 4 \sqrt{5} ,求直线 l_{2} 的方程;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2)当 l_{1}, l_{2} 倾斜角互补时,直线 A C 与直线 B D 交于点 M ,求 \triangle M A B 的内切圆的圆心横坐~\\标的取值范围.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案

(1)l1x,A(2,2),B(2,2),S=12ABxCxD=45l2:y=k(x2),        联立y2=2xxCxD=16k2+4k2=25k2=1k2=15()                       l2:y=±(x2).                                                                                                                  (2)由题意AD,BC关于x轴对称,AC,BD交点在x,AC:y=kx+m,                   联立y2=2x,ky22y+2m=0,y1+y2=2k,y1y2=2mk.                                           kAN+kCN=0,y1x12+y2x22=0,y1y2=4=2mk,m=2k                           AC:y=kx+2k,AC(2,0)AC,BD交点M(2,0),                                         MA,MD关于x轴对称,MAB的圆心Qx轴上,                                                       Q(t,0),A(x1,y1)MQNQ=MANAt+2t2=(x1+2)2+y12(x12)2+y12=x12+6x1+4x122x1+4=1+8x1x122x1+4=1+8x1+4x12(1,5)t(0,35).                           (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, \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}{t - 2}=\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}).~~~~~~~~~~~~~~~~~~~~~~~~~~~

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