圆锥曲线综合题-基础篇

1.椭圆C:x2a2+y2b2=1(a>b>0),𝐹为左焦点,上顶点𝑃𝐹的距离为2,且离心率    32.(1)求椭圆𝐶的标准方程;(2)设斜率为𝑘的动直线𝑙与椭圆𝐶交于𝑀,𝑁两点,且   𝑃𝑀=𝑃𝑁,𝑘的取值范围.                                                                                              1.椭圆C:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0),𝐹为左焦点,上顶点𝑃到𝐹的距离为2,且离心率~~~~\\为\frac{\sqrt{3}}{2}. (1)求椭圆𝐶的标准方程; (2)设斜率为𝑘的动直线𝑙与椭圆𝐶交于𝑀,𝑁两点,且~~~\\|𝑃𝑀|=|𝑃𝑁|,求𝑘的取值范围.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

2.A,B,C是我方三个炮兵阵地,AB正东6kmCB北偏西300,相距4kmP为        炮兵阵地.某时刻A处发现敌炮阵地的某种信号,由于BC两地比AP地远,因此4s    ,BC才同时发现这一信号,此信号的传播速度为1 km/s,A炮击P地,求炮击的    方向角。                                                                                                                2.A,B,C 是我方三个炮兵阵地,A在 B 正东 6km,C 在 B 北偏西 30^0,相距 4 km,P为~~~~~~~~\\炮兵阵地. 某时刻 A 处发现敌炮阵地的某种信号,由于 BC两地比 A距 P 地远,因此 4 s~~~~\\后 ,BC 才同时 发现这一信号,此信号的传播速度为 1 km/s,若 A炮击P地,求炮击的~~~~\\方向角。~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\hspace{1.8cm}

3.已知椭圆𝐶:𝑥2𝑎2+𝑦2𝑏2=1(𝑎>𝑏>0)的短轴长为23,左顶点𝐴到右焦点𝐹的距离为3.(1)求椭圆𝐶的方程;                                                                                                                   (2)若直线𝑙与椭圆𝐶交于不同的两点𝑀,𝑁(不同于𝐴),且直线𝐴𝑀𝐴𝑁的斜率之积与   椭圆的离心率互为相反数,求证:直线𝑙过定点.                                                                    3.已知椭圆𝐶:\frac{𝑥_2}{𝑎_2}+\frac{𝑦_2}{𝑏_2}=1(𝑎>𝑏>0)的短轴长为2\sqrt{3},左顶点𝐴到右焦点𝐹的距离为3.\\ (1)求椭圆𝐶的方程;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2)若直线𝑙与椭圆𝐶交于不同的两点𝑀,𝑁(不同于𝐴),且直线𝐴𝑀和𝐴𝑁的斜率之积与~~~\\椭圆的离心率互为相反数,求证:直线𝑙过定点.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
练习1. 设抛物线C:y2=2px(p>0)的焦点为F,已知点F到圆E:(x+3)2+y2=1     上一点的距离的最大值为6.                                                                                                     (1)求抛物线C的方程.                                                                                                               (2)O是坐标原点,P(2,4),A,B是抛物线C上异于点P的两点,直线PA,PBy轴     分别相交于M,N两点(异于点O),O是线段MN的中点,试判断直线AB是否经过定    .若是,求出该定点坐标;若不是,说明理由.                                                                      练习1. 设抛物线C:y^{2}=2px(p > 0)的焦点为F,已知点F到圆E:(x + 3)^{2}+y^{2}=1~~~~~\\上一点的距离的最大值为6.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (1)求抛物线C的方程.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2)设O是坐标原点,点P(2,4),A,B是抛物线C上异于点P的两点,直线PA,PB与y轴~~~~~\\分别相交于M,N两点(异于点O),且O是线段MN的中点,试判断直线AB是否经过定~~~~\\点.若是,求出该定点坐标;若不是,说明理由.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:(1)y2=8x,(2)过定点(0,2);                                                                                     (2)AB:x=my+nA(x1,y1)B(x2,y2),与抛物线联立得:{x=my+ny2=8x,          y28my8n=0,Δ=64m2+32n>0,y1+y2=8my1y2=8n.                        直线PA:y4=k1(x2)x=0时,yM=2k1+4,                                                   直线PB:y4=k2(x2)x=0时,yN=2k2+4                                                      yM+yN=0k1+k2=4即:y14x12+y24x22=4y14y1282+y24y2282=                 8y1+4+8y2+4=4y1y2+2(y1+y2)=0n=2m.                                                 AB:x=my+2m,过(0,2).                                                                                         答案: (1) y^{2}=8x,(2)过定点(0,-2);~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2) 设 AB:x = my + n,A(x_1,y_1),B(x_2,y_2),与抛物线联立得: \begin{cases}x = my + n\\y^{2}=8x\end{cases},~~~~~~~~~~\\则 y^{2}-8my - 8n = 0, \Delta=64m^{2}+32n>0, y_1 + y_2 = 8m,y_1y_2=-8n.~~~~~~~~~~~~~~~~~~~~~~~~\\ 直线 PA:y - 4 = k_1(x - 2),x = 0 时,y_M=-2k_1 + 4,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 直线 PB:y - 4 = k_2(x - 2),x = 0 时,y_N=-2k_2 + 4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ y_M + y_N = 0\Rightarrow k_1 + k_2 = 4 即:\frac{y_1 - 4}{x_1 - 2}+\frac{y_2 - 4}{x_2 - 2}=4\Rightarrow\frac{y_1 - 4}{\frac{y_1^{2}}{8}-2}+\frac{y_2 - 4}{\frac{y_2^{2}}{8}-2}=~~~~~~~~~~~~~~~~~\\\frac{8}{y_1 + 4}+\frac{8}{y_2 + 4}=4 \Rightarrow y_1y_2+2(y_1 + y_2)=0\Rightarrow n = 2m.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ \therefore AB:x = my + 2m,过 (0,-2).~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

4.已知点B是圆C:(x1)2+y2=16上的任意一点,F(1,0),线段BF的垂直平分线BC于点P.                                                                                                                              (1)求动点P的轨迹E的方程;                                                                                                    (2)设曲线Ex轴的两个交点分别为A1,A2,Q为直线x=4上的动点,Q不在x轴上,    QA1E的另一个交点为M,QA2E的另一个交点为N,证明:FMN的周长为定值.4.已知点 B 是圆 C:(x-1)^{2}+y^{2}=16 上的任意一点,点 F(-1,0) ,线段 B F 的垂直平分线\\交 B C 于点 P .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (1)求动点 P 的轨迹 E 的方程;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2)设曲线 E 与 x 轴的两个交点分别为 A_{1}, A_{2}, Q 为直线 x=4 上的动点,且 Q 不在 x 轴上, ~~~~\\Q A_{1} 与 E 的另一个交点为 M, Q A_{2} 与 E 的另一个交点为 N ,证明: \triangle F M N 的周长为定值.
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答案:(1)x24+y23=1,(2)周长为定值8;                                                                                提示:Q(4,t),直线QA1:y=t6(x+2),直线:QA2:y=t2(x2),                        联立{y=t6(x+2),x24+y23=1,:M(542t227+t2,18t27+t2),同理得:N(2t263+t2,6t3+t2),     直线MN:y+6t3+t2=6tt29(x2t263+t2),y=6tt29(x1),                      所以直线过定点(1,0),即过传圆的右焦点F2,所以FMN的周长为4a=8.                       kMN不存在,x1=x2=1,周长也等于8.所以FMN的周长为定值8.                              答案:(1)\frac{x^{2}}{4}+\frac{y^{2}}{3}=1 , (2)周长为定值8;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 提示:设 Q(4, t), \Rightarrow 直线Q A_{1}: y=\frac{t}{6}(x+2), 直线:Q A_{2}: y=\frac{t}{2}(x-2) ,~~~~~~~~~~~~~~~~~~~~~~~~\\ 联立 \left\{\begin{array}{l}y=\displaystyle\frac{t}{6}(x+2), \\ \displaystyle\frac{x^{2}}{4}+\frac{y^{2}}{3}=1,\end{array}\right. 得: M\left(\frac{54-2 t^{2}}{27+t^{2}}, \frac{18 t}{27+t^{2}}\right) ,同理得 : N\left(\frac{2 t^{2}-6}{3+t^{2}}, \frac{-6 t}{3+t^{2}}\right) ,~~~~~\\ 直线 M N: y+\frac{6 t}{3+t^{2}}=-\frac{6 t}{t^{2}-9}\left(x-\frac{2 t^{2}-6}{3+t^{2}}\right) ,即 \boxed{y=-\frac{6 t}{t^{2}-9}(x-1)} ,~~~~~~~~~~~~~~~~~~~~~~\\ 所以直线过定点 (1,0) ,即过传圆的右焦点 F_{2} ,所以 \triangle F M N 的周长为 4 a=8 .~~~~~~~~~~~~~~~~~~~~~~~\\当 k_{M N} 不存在, x_{1}=x_{2}=1 ,周长也等于 8 .所以 \triangle F M N 的周长为定值 8 .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

5.已知双曲线C的中心为坐标原点,左焦点为(25,0),离心率为5.                               (1)C的方程;                                                                                                                           (2)C的左,右顶点分别为A1,A2,过点(4,0)的直线与C的左支交于M,N两点,M在   第二象限,直线MA1NA2交于点P,证明:点P在定直线上.                                            5.已知双曲线 C 的中心为坐标原点,左焦点为 (-2 \sqrt{5}, 0) ,离心率为 \sqrt{5} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (1)求 C 的方程;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2)记 C 的左,右顶点分别为 A_{1}, A_{2} ,过点 (-4,0) 的直线与 C 的左支交于 M, N 两点, M 在~~~\\第二象限,直线 M A_{1} 与 N A_{2} 交于点 P ,证明:点 P 在定直线上.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:(1)x24y216=1,(2)P在定直线x=1;                                                                  提示:设直线MN的方程为x=my4,联立{x=my4,4x2y2=16.                                      (4m21)y232my+48=0,y1+y2=32m4m21,y1y2=484m21,                      4m210,Δ=256m2+192>0.                                                                                  直线MA1:y=y1x1+2(x+2),直线NA2:y=y2x22(x2).                                           联立得:x+2x2=y2(x1+2)y1(x22)=my1y22(y1+y2)+2y1my1y26y1=16m4m21+2y148m4m216y1          =13,所以x=1,即点P在定直线x=1.                                                                   答案:(1)\frac{x^{2}}{4}-\frac{y^{2}}{16}=1 , (2)P 在定直线 x=-1 上;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 提示:设直线 M N 的方程为 x=m y-4 , 联立 \left\{\begin{array}{l}x=m y-4, \\ 4 x^{2}-y^{2}=16 .\end{array}\right. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\得 \left(4 m^{2}-1\right) y^{2}-32 m y+48=0 , 则 y_{1}+y_{2}=\frac{32 m}{4 m^{2}-1}, y_{1} y_{2}=\frac{48}{4 m^{2}-1} ,~~~~~~~~~~~~~~~~~~~~~~\\且 4 m^{2}-1 \neq 0, \Delta=256 m^{2}+192>0 .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 直线 M A_{1} : y=\frac{y_{1}}{x_{1}+2}(x+2) ,直线 N A_{2} : y=\frac{y_{2}}{x_{2}-2}(x-2) . ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\联立得:\boxed{ \frac{x+2}{x-2}=\frac{y_{2}\left(x_{1}+2\right)}{y_{1}\left(x_{2}-2\right)}=\frac{m y_{1} y_{2}-2\left(y_{1}+y_{2}\right)+2 y_{1}}{m y_{1} y_{2}-6 y_{1}}}=\frac{-\displaystyle\frac{16 m}{4 m^{2}-1}+2 y_{1}}{\displaystyle\frac{48 m}{4 m^{2}-1}-6 y_{1}}~~~~~~~~~~\\=-\frac{1}{3} , 所以 x=-1 ,即点 P 在定直线 x=-1 上.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

6.已知O为坐标原点,动直线l:y=kx+m(km0)与双曲线C:x2y2b2=1(b>0)渐近线交于A,B两点,与椭圆D:x22+y2=1交于E,F两点.k2=10,2(OA+OB)=3(OE+OF).                                                                                                                         (1)求双曲线C的方程;(2)若动直线l与双曲线C相切,求证:OAB的面积为定值.         6.已知 O 为坐标原点,动直线 l: y=k x+m(k m \neq 0) 与双曲线 C: x^{2}-\frac{y^{2}}{b^{2}}=1(b>0) 的\\渐近线交于 A, B 两点,与椭圆 D: \frac{x^{2}}{2}+y^{2}=1 交于 E, F 两点.当 k^{2}=10 时, 2(\overrightarrow{O A} +\overrightarrow{O B})\\=3(\overrightarrow{O E}+\overrightarrow{O F}) .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (1)求双曲线 C 的方程; (2)若动直线 l 与双曲线 C 相切,求证: \triangle O A B 的面积为定值.~~~~~~~~~
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答案:(1)x2y23=1,(2)3;                                                                                                  提示:(1)2(OA+OB)=3(OE+OF),所以x1+x2=32(x3+x4).{y=kx+m,y=bx,x1=mbk同理可得x2=mb+k.{y=kx+m,x22+y2=1,                                                    (1+2k2)x2+4kmx+2m22=0,所以x3+x4=4km1+2k2.                                    解得b=3.所以双曲线C的方程为x2y23=1.                                                                  (2)证明双曲线的渐近线方程为y=±3x.(1)A(mk3,3mk3),                      B(mk+3,3mk+3),AOB=2π3.                                                                                 法一,所以SOAB=12OAOBsinAOB=                                                                 124m2k2332=3m2k23.{y=kx+m,x2y23=1,(3k2)x22kmxm23=0.所以Δ=4k2m2+4(3k2)(m2+3)=0,m2=k23,所以SOAB=3m2k23=3.法二、原点OAB距离d=mk2+1,AB=1+k2xAxB=                               1+k2mk3+mk+3,SAOB=12ABd=3m2k23=3                               法三、设曲线C上一点为(x0,y0)x02y023=1切线:x0xy03y=1y=3x联立,xA=33x03y0,yA=333x03y0,y=3x联立,xB=33x0+3y0,      yB=333x0+3y0.y=0切线与x轴交点(1x0,0).                                                    SAOB=121x0yByA=121x03313x03y0+13x0+3y0=3321x06x09x023y02,9x023y02=9=3答案:(1)x^{2}-\frac{y^{2}}{3}=1 ,(2)\sqrt{3};~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 提示:(1)2(\overrightarrow{O A}+\overrightarrow{O B})=3(\overrightarrow{O E}+\overrightarrow{O F}) ,所以 x_{1}+x_{2}=\frac{3}{2}\left(x_{3}+x_{4}\right) . 由 \left\{\begin{array}{l}y=k x+m, \\ y=b x,\end{array}\right. \\得 x_{1}=\frac{m}{b-k} 同理可得 x_{2}=-\frac{m}{b+k} . 由 \left\{\begin{array}{l}y=k x+m, \\ \frac{x^{2}}{2}+y^{2}=1,\end{array}\right. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\得 \left(1+2 k^{2}\right) x^{2}+4 k m x+2 m^{2}-2=0 ,所以 x_{3}+x_{4}=-\frac{4 k m}{1+2 k^{2}} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 解得 b=\sqrt{3} .所以双曲线 C 的方程为 x^{2}-\frac{y^{2}}{3}=1 .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2)证明 双曲线的渐近线方程为 y= \pm \sqrt{3} x . 由(1)得 A\left(\frac{-m}{k-\sqrt{3}}, \frac{-\sqrt{3} m}{k-\sqrt{3}}\right), ~~~~~~~~~~~~~~~~~~~~~~\\ B\left(\frac{-m}{k+\sqrt{3}}, \frac{\sqrt{3} m}{k+\sqrt{3}}\right), \angle A O B=\frac{2 \pi}{3} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 法一,所以 \boxed{ S_{\triangle O A B}=\frac{1}{2}|O A| \cdot|O B| \cdot \sin \angle A O B}=~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ \frac{1}{2} \cdot\left|\frac{4 m^{2}}{k^{2}-3}\right| \cdot \frac{\sqrt{3}}{2}=\left|\frac{\sqrt{3} m^{2}}{k^{2}-3}\right| . 由 \left\{\begin{array}{l}y=k x+m, \\ x^{2}-\displaystyle\frac{y^{2}}{3}=1,\end{array}\right. 得 \left(3-k^{2}\right) x^{2}-2 k m x-m^{2}-3=0 .\\ 所以 \Delta=4 k^{2} m^{2}+4\left(3-k^{2}\right)\left(m^{2}+3\right)=0 ,得 m^{2}=k^{2}-3 , 所以 S_{O A B}=\left|\frac{\sqrt{3} m^{2}}{k^{2}-3}\right|=\sqrt{3} .\\ 法二、原点 O 到 AB 距离 d = \frac{|m|}{\sqrt{k^{2}+1}},|AB|=\sqrt{1 + k^{2}}|x_{A}-x_{B}|=~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\\sqrt{1 + k^{2}}|\frac{-m}{k - \sqrt{3}}+\frac{m}{k+\sqrt{3}}|, \boxed{S_{\triangle AOB}=\frac{1}{2}|AB|\cdot d}=\left|\frac{\sqrt{3}m^{2}}{k^{2}-3}\right|=\sqrt{3}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 法三、设曲线 C 上一点为 \left(x_{0}, y_{0}\right) \therefore x_{0}^{2}-\frac{y_{0}^{2}}{3}=1 切线: x_{0} x-\frac{y_{0}}{3} y=1 与 y=\sqrt{3} x 联立,\\ \Rightarrow x_{A}=\frac{3}{3 x_{0}-\sqrt{3} y_{0}}, y_{A}= \frac{3 \sqrt{3}}{3 x_{0}-\sqrt{3} y_{0}}, 与 y=-\sqrt{3} x 联立, \Rightarrow x_{B}=\frac{3}{3 x_{0}+\sqrt{3} y_{0}}, ~~~~~~\\y_{B} =\frac{-3 \sqrt{3}}{3 x_{0}+\sqrt{3} y_{0}}. 令 y=0 \Rightarrow 切线与 x 轴交点 \left(\frac{1}{x_{0}}, 0\right) .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ \begin{aligned} \therefore S_{\triangle A O B} & =\frac{1}{2}\left|\frac{1}{x_{0}}\right|\left|y_{B}-y_{A}\right| \\ & =\frac{1}{2}\left|\frac{1}{x_{0}}\right| \cdot 3 \sqrt{3}\left|\frac{1}{3 x_{0}-\sqrt{3} y_{0}}+\frac{1}{3 x_{0}+\sqrt{3} y_{0}}\right| \\ & =\frac{3 \sqrt{3}}{2}\left|\frac{1}{x_{0}}\right|\left|\frac{6 x_{0}}{9 x_{0}^{2}-3 y_{0}^{2}}\right|, \quad 9 x_{0}^{2}-3 y_{0}^{2}=9 \\ & =\sqrt{3} \end{aligned}

7.已知椭圆C:x2a2+y2b2=1(a>b>0)经过点(2,3),离心率为32.                                (1)求椭圆C的标准方程;(2)若直线l:y=kx+m与椭圆C有两个不同的交点A,B,原   O到直线l的距离为2,ABO的面积的最大值.                                                              7.已知椭圆 C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0) 经过点 (2, \sqrt{3}) ,离心率为 \frac{\sqrt{3}}{2} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (1)求椭圆 C 的标准方程; (2)若直线 l: y=k x+m 与椭圆 C 有两个不同的交点 A, B ,原~~~\\点 O 到直线 l 的距离为 2 ,求 \triangle A B O 的面积的最大值.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:(1)x216+y24=1,(2)4;(2)提示:原点O到直线l的距离为2,m2=4(k2+1).{y=kx+m,x216+y24=1,(1+4k2)x2+8kmx+4m216=0,                                          Δ=16(16k2+4m2)=192k2>0                        ★使用公式AB=1+k2ΔaAB=83k1+k21+4k2.所以ABO面积S=83k1+k21+4k28×3k2+1+k221+4k2 =4,当且仅当k=±22时取得等号,所以ABO的面积的最大值为4.                            答案:解(1) \frac{x^{2}}{16}+\frac{y^{2}}{4}=1,(2)4; (2)提示:原点 O 到直线 l 的距离为 2, \Rightarrow m^{2}=4\left(k^{2}+1\right) .\\ 由 \left\{\begin{array}{l}y=k x+m, \\ \displaystyle\frac{x^{2}}{16}+\frac{y^{2}}{4}=1,\end{array}\right. 得 \left(1+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-16=0 ,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ \Delta=16\left(16 k^{2}+4-m^{2}\right)=192 k^{2}>0~~~~~~~~~~~~~~~~~~~~~~~~ ★ 使用公式 |A B|=\sqrt{1+k^{2}} \frac{\sqrt{\Delta}}{|a|}★ \\ 则 \boxed{ |A B|=\frac{8 \sqrt{3}|k| \sqrt{1+k^{2}}}{1+4 k^{2}} }. 所以 \triangle A B O 面积 S=\frac{8 \sqrt{3}|k| \sqrt{1+k^{2}}}{1+4 k^{2}}\leqslant \frac{8 \times \displaystyle\frac{3 k^{2}+1+k^{2}}{2}}{1+4 k^{2}}~\\=4 ,当且仅当 k= \pm \frac{\sqrt{2}}{2} 时取得等号, 所以 \triangle A B O 的面积的最大值为 4 .~~~~~~~~~~~~~~~~~~~~~~~~~~~~

8.如图,已知抛物线x2=y,A(12,14),B(32,94),拋物线上的点P(x,y),其中         12<x<32,过点B作直线AP的垂线,垂足为Q.求PAPQ的最大值.                8.如图,已知抛物线 x^{2}=y ,点 A\left(-\frac{1}{2}, \frac{1}{4}\right), B\left(\frac{3}{2}, \frac{9}{4}\right),拋物线上的点 P(x, y),其中~~~~~~~~~\\-\frac{1}{2} < x< \frac{3}{2}, 过点 B 作直线 A P 的垂线,垂足为 Q .求 |P A| \cdot|P Q| 的最大值.~~~~~~~~~~~~~~~~
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答案:2716;提示:P(x,y),因为y=x2.由已知得PAPB=PAPB                    =(x+12,y14)(x32,y94)=[(x+12)(x32)+(x214)(x294)]              =x4+32x2+x+316=f(x),f(x)=4x3+3x+1=3x3x3+1x3             =(1x)(2x+1)2x=1为极大值点.所以f(x)max=2716.                                                答案:\frac{27}{16};提示:设 P(x,y),因为 y = x^{2}.由已知得 \boxed{\vert PA\vert\vert PB\vert=-\overrightarrow{PA}\cdot\overrightarrow{PB}}~~~~~~~~~~~~~~~~~~~~\\=-(x + \frac{1}{2},y-\frac{1}{4})\cdot(x-\frac{3}{2},y - \frac{9}{4}) =-\left[(x+\frac{1}{2})(x - \frac{3}{2})+(x^{2}-\frac{1}{4})(x^{2}-\frac{9}{4})\right]~~~~~~~~~~~~~~\\ =-x^{4}+\frac{3}{2}x^{2}+x+\frac{3}{16}=f(x),\Rightarrow f^\prime(x)=-4x^{3}+3x + 1=3x-3x^{3}+1 - x^{3}~~~~~~~~~~~~~\\=(1 - x)(2x + 1)^{2},x = 1为极大值点. 所以 f(x)_{\max}=\frac{27}{16} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

9.已知点A(m,4)(m>0)在抛物线x2=4y上,过点A作倾斜角互补的两条直线l1l2l1,l2与抛物线的另一个交点分别为B,C.                                                                            (1)求证:直线BC的斜率为定值;                                                                                         (2)若拋物线上存在两点关于BC对称,求BC的取值范围.                                             9.已知点 A(m, 4)(m>0) 在抛物线 x^{2}=4 y 上,过点 A 作倾斜角互补的两条直线 l_{1} 和 l_{2} 且\\ l_{1}, l_{2} 与抛物线的另一个交点分别为 B, C .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (1)求证:直线 B C 的斜率为定值;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2)若拋物线上存在两点关于 B C 对称,求 |B C| 的取值范围.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案(1)2,(2)105;提示:{y4=k(x4)x2=4yx24kx+16k16=0,x1x2=16k16xB=4k4,同理xC=4k4,kBC=yByCxBxC=14(xB+xC)=2.     (2)BCy=2x+t,BC中垂线:y=12x+m,{y=12x+mx2=4yx22x4m=0,Δ=4+16m>0m>14,x1+x2=2=2x0,得x0=1y0=12+m,代入BCm=52+t>14t>94.{y=2x+tx2=4yx2+8x4t=0,BC=                 1+k2Δa=564+16t>105                                                                                答案(1)-2,(2)10\sqrt{5};提示:由\begin{cases}y - 4=k(x - 4)\\x^{2}=4y\end{cases} \Rightarrow x^{2}-4kx + 16k-16 = 0, x_{1}x_{2}=\\16k - 16\Rightarrow x_{B}=4k - 4,同理x_{C}=-4k - 4, k_{BC}=\frac{y_{B}-y_{C}}{x_{B}-x_{C}}=\frac{1}{4}(x_{B}+x_{C})=-2.~~~~~\\ (2) BC:y=-2x + t, BC中垂线:y=\frac{1}{2}x + m, 由\begin{cases}y=\displaystyle\frac{1}{2}x + m\\x^{2}=4y\end{cases}\Rightarrow x^{2}-2x-4m = 0,\\ \boxed{\Delta=4 + 16m>0,m>-\frac{1}{4},} x_{1}+x_{2}=2 = 2x_{0},得x_{0}=1,y_{0}=\frac{1}{2}+m, 代入BC得\\\boxed{m=-\frac{5}{2}+t>-\frac{1}{4}\Rightarrow t>\frac{9}{4}.} 由\begin{cases}y=-2x + t\\x^{2}=4y\end{cases}\Rightarrow x^{2}+8x-4t = 0, \vert BC\vert= ~~~~~~~~~~~~~~~~~\\\boxed{\sqrt{1 + k^{2}}\frac{\sqrt{\Delta}}{\vert a\vert}}=\sqrt{5}\sqrt{64 + 16t}>10\sqrt{5}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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