Question

7£®ÍÖÔ²C£º$\frac{{x}^{2}}{{a}^{2}}$+$\frac{{y}^{2}}{{b}^{2}}$=1£¨a£¾b£¾0£©µÄÉ϶¥µãΪA£¬P£¨$\frac{4\sqrt{2}}{3}$£¬$\frac{b}{3}$£©ÊÇCÉϵÄÒ»µã£¬ÒÔAPΪֱ¾¶µÄÔ²¾­¹ýÍÖÔ²CµÄÓÒ½¹µãF£®
£¨1£©ÇóÍÖÔ²CµÄ·½³Ì£»
£¨2£©ÈôÖ±Ïßl£ºy=kx+m£¨|k|¡Ü$\frac{\sqrt{2}}{2}$£©ÓëÍÖÔ²CÏཻÓÚA¡¢BÁ½µã£¬MΪÍÖÔ²CÉÏÈÎÒâÒ»µã£¬ÇÒÏ߶ÎOMµÄÖеãÓëÏ߶ÎABµÄÖеãÖغϣ¬Çó|OM|µÄÈ¡Öµ·¶Î§£®

Answer 1

·ÖÎö £¨1£©ÓÉÓÚÒÔAPΪֱ¾¶µÄÔ²¾­¹ýÍÖÔ²CµÄÓÒ½¹µãF£¬¿ÉµÃ$\overrightarrow{PF}•\overrightarrow{AF}$=0£¬ÔÙÓɵãP£¨$\frac{4\sqrt{2}}{3}$£¬$\frac{b}{3}$£©ÔÚÍÖÔ²ÉÏ£¬ÁªÁ¢¿ÉµÃa£¬b£¬cµÄÖµ£¬ÔòÍÖÔ²·½³Ì¿ÉÇó£»
£¨2£©ÁªÁ¢Ö±Ïß·½³ÌºÍÍÖÔ²·½³Ì£¬Çó³öA£¬BÖеã×ø±ê£¬µÃµ½M×ø±ê£¬°ÑM×ø±ê´úÈëÍÖÔ²·½³Ì£¬¿ÉµÃmÓëkµÄ¹Øϵ£¬°Ñ|OM|»¯Îªº¬ÓÐkµÄ´úÊýʽ£¬½áºÏÒÑÖªkµÄ·¶Î§ÇóµÃ|OM|µÄÈ¡Öµ·¶Î§£®

½â´ð ½â£º£¨1£©A£¨0£¬b£©£®
¡ßÒÔAPΪֱ¾¶µÄÔ²¾­¹ýÍÖÔ²CµÄÓÒ½¹µãF£¬¡àPF¡ÍAF£¬
¡à$\overrightarrow{PF}•\overrightarrow{AF}$=£¨c-$\frac{4\sqrt{2}}{3}$£¬-$\frac{b}{3}$£©•£¨c£¬-b£©=c£¨c-$\frac{4\sqrt{2}}{3}$£©+$\frac{{b}^{2}}{3}$=0£®
°ÑµãP£¨$\frac{4\sqrt{2}}{3}$£¬$\frac{b}{3}$£©´úÈëÍÖÔ²C£º$\frac{{x}^{2}}{{a}^{2}}$+$\frac{{y}^{2}}{{b}^{2}}$=1£¬
µÃ$\frac{32}{9{a}^{2}}+\frac{1}{9}=1$£¬½âµÃa²=4£¬
¡àb²+c²=4£¬¿ÉµÃb²=4-c²£¬´úÈëc£¨c-$\frac{4\sqrt{2}}{3}$£©+$\frac{{b}^{2}}{3}=0$£¬½âµÃc=$\sqrt{2}$£¬b=$\sqrt{2}$£®
¡àÍÖÔ²CµÄ·½³ÌΪ$\frac{{x}^{2}}{4}+\frac{{y}^{2}}{2}=1$£»
£¨2£©Èçͼ£¬ÉèA£¨x₁£¬y₁£©£¬B£¨x₂£¬y₂£©£¬
ÁªÁ¢$\left\{\begin{array}{l}{y=kx+m}\\{\frac{{x}^{2}}{4}+\frac{{y}^{2}}{2}=1}\end{array}\right.$£¬µÃ£¨1+2k²£©x²+4kmx+2m²-4=0£®
¡÷=£¨4km£©²-4£¨1+2k²£©£¨2m²-4£©=32k²-8m²+16£¾0£¬¼´4k²-m²+2£¾0  ¢Ù£®
${x}_{1}+{x}_{2}=\frac{-4km}{1+2{k}^{2}}$£¬
¡à${y}_{1}+{y}_{2}=k£¨{x}_{1}+{x}_{2}£©+2m=k•\frac{-4km}{1+2{k}^{2}}+2m$=$\frac{2m}{1+2{k}^{2}}$£®
¡àABÖеãG£¨$-\frac{2km}{1+2{k}^{2}}£¬\frac{m}{1+2{k}^{2}}$£©£¬
ÔòM£¨$-\frac{4km}{1+2{k}^{2}}£¬\frac{2m}{1+2{k}^{2}}$£©£¬
¡ßMÔÚÍÖÔ²$\frac{{x}^{2}}{4}+\frac{{y}^{2}}{2}=1$ÉÏ£¬
¡à$\frac{4{k}^{2}{m}^{2}}{£¨1+2{k}^{2}£©^{2}}+\frac{2{m}^{2}}{£¨1+2{k}^{2}£©^{2}}=1$£¬ÕûÀíµÃ£º${m}^{2}=\frac{1+2{k}^{2}}{2}$£®
°Ñ${m}^{2}=\frac{1+2{k}^{2}}{2}$´úÈë¢ÙµÃ£¬$4{k}^{2}-\frac{1+2{k}^{2}}{2}+2£¾0$ºã³ÉÁ¢£®
¡à|OM|=$\sqrt{£¨-\frac{4km}{1+2{k}^{2}}£©^{2}+£¨\frac{2m}{1+2{k}^{2}}£©^{2}}$=$\sqrt{\frac{16{k}^{2}}{£¨1+2{k}^{2}£©^{2}}•\frac{1+2{k}^{2}}{2}+\frac{4}{£¨1+2{k}^{2}£©^{2}}•\frac{1+2{k}^{2}}{2}}$
=$\sqrt{\frac{8{k}^{2}+2}{2{k}^{2}+1}}$=$\sqrt{\frac{4£¨2{k}^{2}+1£©-2}{2{k}^{2}+1}}=\sqrt{4-\frac{2}{2{k}^{2}+1}}$£®
¡ß|k|¡Ü$\frac{\sqrt{2}}{2}$£¬¡à1¡Ü2k²+1¡Ü2£¬Ôò$4-\frac{2}{2{k}^{2}+1}¡Ê[2£¬3]$£¬
¡à|OM|µÄÈ¡Öµ·¶Î§Îª$[\sqrt{2}£¬\sqrt{3}]$£®

µãÆÀ ±¾Ì⿼²éÁËÍÖÔ²µÄ±ê×¼·½³Ì¼°ÆäÐÔÖÊ¡¢Ö±ÏßÓëÍÖÔ²ÏཻÏÒ³¤ÎÊÌ⣬¿¼²éÁËÍÆÀíÂÛÖ¤ÄÜÁ¦Óë¼ÆËãÄÜÁ¦£¬ÌåÏÖÁËÕûÌåÔËËã˼Ïë·½·¨£¬ÊôÓÚÄÑÌ⣮