Question

7£®ÒÑÖªµãA£¬BÊÇÍÖÔ²C£º$\frac{x^2}{a^2}+\frac{y^2}{b^2}$=1£¨a£¾b£¾0£©µÄ×ó¡¢ÓÒ¶¥µã£¬FΪ×󽹵㣬µãPÊÇÍÖÔ²ÉÏÒìÓÚA£¬BµÄÈÎÒâÒ»µã£¬Ö±ÏßAPÓë¹ýµãBÇÒ´¹Ö±ÓÚxÖáµÄÖ±Ïßl½»ÓÚµãM£¬Ö±ÏßMN¡ÍBPÓÚµãN£®
£¨1£©ÇóÖ¤£ºÖ±ÏßAPÓëÖ±ÏßBPµÄбÂÊÖ®»ýΪ¶¨Öµ£»
£¨2£©ÈôÖ±ÏßMN¹ý½¹µãF£¬$\overrightarrow{AF}=¦Ë\overrightarrow{FB}$£¨¦Ë¡ÊR£©£¬ÇóʵÊý¦ËµÄÖµ£®

Answer 1

·ÖÎö £¨1£©¸ù¾ÝÌâÒ⣬ÉèP£¨x₀£¬y₀£©£¬ÓÉPµÄ×ø±ê±íʾֱÏßAPÓëÖ±ÏßBPµÄбÂÊ£¬ÇóÆä»ý¿ÉµÃ${k_{AP}}•{k_{BP}}=\frac{y_0}{{{x_0}+a}}•\frac{y_0}{{{x_0}-a}}=\frac{{{y_0}^2}}{{{x_0}^2-{a^2}}}$£¬ÓÉÍÖÔ²µÄÐÔÖʼ´¿ÉµÃÖ¤Ã÷£»
£¨2£©ÉèÖ±ÏßAPÓëBPбÂÊ·Ö±ðΪk₁¡¢k₂£¬½ø¶ø¿ÉµÃÖ±ÏßAPµÄ·½³Ì£¬·ÖÎö¿ÉµÃ${k_{MN}}=-\frac{a^2}{b^2}•{k_1}$£¬ÓÖF¡¢N¡¢MÈýµã¹²Ïߣ¬µÃk_MF=k_MN£¬¼´$\frac{{2a{k_1}}}{a+c}=\frac{a^2}{b^2}•{k_1}$£¬ÓÉÏòÁ¿µÄÊý³ËÔËËãµÄÒâÒå·ÖÎö¿ÉµÃÖ¤Ã÷£®

½â´ð ½â£º£¨1£©Ö¤Ã÷£ºÉèP£¨x₀£¬y₀£©£¨x₀¡Ù¡Àa£©£¬
ÓÉÒÑÖªA£¨-a£¬0£©£¬B£¨a£¬0£©£¬
¡à${k_{AP}}•{k_{BP}}=\frac{y_0}{{{x_0}+a}}•\frac{y_0}{{{x_0}-a}}=\frac{{{y_0}^2}}{{{x_0}^2-{a^2}}}$£®¢Ù
¡ßµãPÔÚÍÖÔ²ÉÏ£¬¡à$\frac{{{x_0}^2}}{a^2}+\frac{{{y_0}^2}}{b^2}=1$£®¢Ú
ÓÉ¢Ù¢ÚµÃ${k_{AP}}•{k_{BP}}=\frac{{{y_0}^2}}{{{x_0}^2-{a^2}}}=\frac{{-\frac{b^2}{a^2}£¨{x_0}^2-{a^2}£©}}{{{x_0}^2-{a^2}}}=-\frac{b^2}{a^2}$£¨¶¨Öµ£©£®
¡àÖ±ÏßAPÓëÖ±ÏßBPµÄбÂÊÖ®»ýΪ¶¨Öµ$-\frac{b^2}{a^2}$£®
£¨2£©ÉèÖ±ÏßAPÓëBPбÂÊ·Ö±ðΪk₁¡¢k₂£¬
ÓÉÒÑÖªF£¨-c£¬0£©£¬Ö±ÏßAPµÄ·½³ÌΪy=k₁£¨x+a£©£¬
Ö±Ïßl£ºx=a£¬ÔòM£¨a£¬2ak₁£©£®
¡ßMN¡ÍBP£¬¡àk_MN•k₂=-1£®
ÓÉ£¨1£©Öª${k_1}•{k_2}=-\frac{b^2}{a^2}$£¬¹Ê${k_{MN}}=-\frac{a^2}{b^2}•{k_1}$£¬
ÓÖF¡¢N¡¢MÈýµã¹²Ïߣ¬
µÃk_MF=k_MN£¬¼´$\frac{{2a{k_1}}}{a+c}=\frac{a^2}{b^2}•{k_1}$£¬
µÃ2b²=a£¨a+c£©£®
¡ßb²=a²-c²£¬
¡à2£¨a²-c²£©=a²+ac£¬2c²+ac-a²=0£¬$2{£¨\frac{c}{a}£©^2}+\frac{c}{a}-1=0$£¬
½âµÃ$\frac{c}{a}=\frac{1}{2}$»ò$\frac{c}{a}=-1$£¨ÉáÈ¥£©£®
¡àa=2c£®
ÓÉÒÑÖª$\overrightarrow{AF}=¦Ë\overrightarrow{FB}$£¬µÃ£¨a-c£¬0£©=¦Ë£¨a+c£¬0£©£¬
½«a=2c´úÈ룬µÃ£¨c£¬0£©=¦Ë£¨3c£¬0£©£¬¹Ê$¦Ë=\frac{1}{3}$£®

µãÆÀ ±¾Ì⿼²éÖ±ÏßÓëÍÖÔ²µÄλÖùØϵ£¬Éæ¼°ÍÖÔ²µÄ¼¸ºÎÐÔÖÊ£¬¹Ø¼üÒªÊìϤÍÖÔ²µÄ¼¸ºÎÐÔÖÊ£®