---

---

参考图象

答案

$$答案: (1) y^{2}=8x,(2)过定点(0,-2);~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2) 设 AB:x = my + n，A(x_1,y_1)，B(x_2,y_2),与抛物线联立得: \begin{cases}x = my + n\\y^{2}=8x\end{cases},~~~~~~~~~~\\则 y^{2}-8my - 8n = 0, \Delta=64m^{2}+32n>0, y_1 + y_2 = 8m，y_1y_2=-8n.~~~~~~~~~~~~~~~~~~~~~~~~\\ 直线 PA:y - 4 = k_1(x - 2)，x = 0 时，y_M=-2k_1 + 4,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 直线 PB:y - 4 = k_2(x - 2)，x = 0 时，y_N=-2k_2 + 4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ y_M + y_N = 0\Rightarrow k_1 + k_2 = 4 即：\frac{y_1 - 4}{x_1 - 2}+\frac{y_2 - 4}{x_2 - 2}=4\Rightarrow\frac{y_1 - 4}{\frac{y_1^{2}}{8}-2}+\frac{y_2 - 4}{\frac{y_2^{2}}{8}-2}=~~~~~~~~~~~~~~~~~\\\frac{8}{y_1 + 4}+\frac{8}{y_2 + 4}=4 \Rightarrow y_1y_2+2(y_1 + y_2)=0\Rightarrow n = 2m.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ \therefore AB:x = my + 2m，过 (0,-2).~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

---

参考图象

答案

$$答案:(1)\frac{x^{2}}{4}+\frac{y^{2}}{3}=1 , (2)周长为定值8;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 提示:设 Q(4, t), \Rightarrow 直线Q A_{1}: y=\frac{t}{6}(x+2), 直线:Q A_{2}: y=\frac{t}{2}(x-2) ,~~~~~~~~~~~~~~~~~~~~~~~~\\ 联立 \left\{\begin{array}{l}y=\displaystyle\frac{t}{6}(x+2), \\ \displaystyle\frac{x^{2}}{4}+\frac{y^{2}}{3}=1,\end{array}\right. 得: M\left(\frac{54-2 t^{2}}{27+t^{2}}, \frac{18 t}{27+t^{2}}\right) ,同理得 : N\left(\frac{2 t^{2}-6}{3+t^{2}}, \frac{-6 t}{3+t^{2}}\right) ,~~~~~\\ 直线 M N: y+\frac{6 t}{3+t^{2}}=-\frac{6 t}{t^{2}-9}\left(x-\frac{2 t^{2}-6}{3+t^{2}}\right) ,即 \boxed{y=-\frac{6 t}{t^{2}-9}(x-1)} ,~~~~~~~~~~~~~~~~~~~~~~\\ 所以直线过定点 (1,0) ,即过传圆的右焦点 F_{2} ,所以 \triangle F M N 的周长为 4 a=8 .~~~~~~~~~~~~~~~~~~~~~~~\\当 k_{M N} 不存在, x_{1}=x_{2}=1 ,周长也等于 8 .所以 \triangle F M N 的周长为定值 8 .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

---

参考图象

答案

$$答案:(1)\frac{x^{2}}{4}-\frac{y^{2}}{16}=1 , (2)P 在定直线 x=-1 上;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 提示:设直线 M N 的方程为 x=m y-4 , 联立 \left\{\begin{array}{l}x=m y-4, \\ 4 x^{2}-y^{2}=16 .\end{array}\right. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\得 \left(4 m^{2}-1\right) y^{2}-32 m y+48=0 , 则 y_{1}+y_{2}=\frac{32 m}{4 m^{2}-1}, y_{1} y_{2}=\frac{48}{4 m^{2}-1} ,~~~~~~~~~~~~~~~~~~~~~~\\且 4 m^{2}-1 \neq 0, \Delta=256 m^{2}+192>0 .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 直线 M A_{1} : y=\frac{y_{1}}{x_{1}+2}(x+2) ,直线 N A_{2} : y=\frac{y_{2}}{x_{2}-2}(x-2) . ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\联立得：\boxed{ \frac{x+2}{x-2}=\frac{y_{2}\left(x_{1}+2\right)}{y_{1}\left(x_{2}-2\right)}=\frac{m y_{1} y_{2}-2\left(y_{1}+y_{2}\right)+2 y_{1}}{m y_{1} y_{2}-6 y_{1}}}=\frac{-\displaystyle\frac{16 m}{4 m^{2}-1}+2 y_{1}}{\displaystyle\frac{48 m}{4 m^{2}-1}-6 y_{1}}~~~~~~~~~~\\=-\frac{1}{3} , 所以 x=-1 ,即点 P 在定直线 x=-1 上.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

---

参考图象

答案

$$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}$$

---

答案

$$答案:解(1) \frac{x^{2}}{16}+\frac{y^{2}}{4}=1,(2)4; (2)提示:原点 O 到直线 l 的距离为 2, \Rightarrow m^{2}=4\left(k^{2}+1\right) .\\ 由 \left\{\begin{array}{l}y=k x+m, \\ \displaystyle\frac{x^{2}}{16}+\frac{y^{2}}{4}=1,\end{array}\right. 得 \left(1+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-16=0 ,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ \Delta=16\left(16 k^{2}+4-m^{2}\right)=192 k^{2}>0~~~~~~~~~~~~~~~~~~~~~~~~ ★ 使用公式 |A B|=\sqrt{1+k^{2}} \frac{\sqrt{\Delta}}{|a|}★ \\ 则 \boxed{ |A B|=\frac{8 \sqrt{3}|k| \sqrt{1+k^{2}}}{1+4 k^{2}} }. 所以 \triangle A B O 面积 S=\frac{8 \sqrt{3}|k| \sqrt{1+k^{2}}}{1+4 k^{2}}\leqslant \frac{8 \times \displaystyle\frac{3 k^{2}+1+k^{2}}{2}}{1+4 k^{2}}~\\=4 ,当且仅当 k= \pm \frac{\sqrt{2}}{2} 时取得等号, 所以 \triangle A B O 的面积的最大值为 4 .~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

---

答案

$$答案:\frac{27}{16};提示:设 P(x,y)，因为 y = x^{2}.由已知得 \boxed{\vert PA\vert\vert PB\vert=-\overrightarrow{PA}\cdot\overrightarrow{PB}}~~~~~~~~~~~~~~~~~~~~\\=-(x + \frac{1}{2},y-\frac{1}{4})\cdot(x-\frac{3}{2},y - \frac{9}{4}) =-\left[(x+\frac{1}{2})(x - \frac{3}{2})+(x^{2}-\frac{1}{4})(x^{2}-\frac{9}{4})\right]~~~~~~~~~~~~~~\\ =-x^{4}+\frac{3}{2}x^{2}+x+\frac{3}{16}=f(x),\Rightarrow f^\prime(x)=-4x^{3}+3x + 1=3x-3x^{3}+1 - x^{3}~~~~~~~~~~~~~\\=(1 - x)(2x + 1)^{2}，x = 1为极大值点. 所以 f(x)_{\max}=\frac{27}{16} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

---

答案

$$答案(1)-2,(2)10\sqrt{5};提示:由\begin{cases}y - 4=k(x - 4)\\x^{2}=4y\end{cases} \Rightarrow x^{2}-4kx + 16k-16 = 0, x_{1}x_{2}=\\16k - 16\Rightarrow x_{B}=4k - 4,同理x_{C}=-4k - 4, k_{BC}=\frac{y_{B}-y_{C}}{x_{B}-x_{C}}=\frac{1}{4}(x_{B}+x_{C})=-2.~~~~~\\ (2) BC：y=-2x + t, BC中垂线：y=\frac{1}{2}x + m, 由\begin{cases}y=\displaystyle\frac{1}{2}x + m\\x^{2}=4y\end{cases}\Rightarrow x^{2}-2x-4m = 0,\\ \boxed{\Delta=4 + 16m>0，m>-\frac{1}{4},} x_{1}+x_{2}=2 = 2x_{0}，得x_{0}=1，y_{0}=\frac{1}{2}+m， 代入BC得\\\boxed{m=-\frac{5}{2}+t>-\frac{1}{4}\Rightarrow t>\frac{9}{4}.} 由\begin{cases}y=-2x + t\\x^{2}=4y\end{cases}\Rightarrow x^{2}+8x-4t = 0, \vert BC\vert= ~~~~~~~~~~~~~~~~~\\\boxed{\sqrt{1 + k^{2}}\frac{\sqrt{\Delta}}{\vert a\vert}}=\sqrt{5}\sqrt{64 + 16t}>10\sqrt{5}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

---