Question

12£®ÒÑÖªÍÖÔ²$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}$=1£¨a£¾b£¾0£©µÄ×ó¡¢ÓÒ½¹µã·Ö±ðΪF₁£¬F₂£¬|F₁F₂|=2c£¬µãAÔÚÍÖÔ²ÉÏ£¬ÇÒAF₁´¹Ö±ÓÚxÖᣬ$\overrightarrow{A{F}_{1}}$•$\overrightarrow{A{F}_{2}}$=c²£¬ÔòÍÖÔ²µÄÀëÐÄÂÊeµÈÓÚ£¨¡¡¡¡£©

• A£® $\frac{\sqrt{3}}{3}$
• B£® $\frac{\sqrt{3}-1}{2}$
• C£® $\frac{\sqrt{5}-1}{2}$
• D£® $\frac{\sqrt{2}}{2}$

Answer 1

·ÖÎö ÓÉÌâÒâ»­³öͼÐΣ¬µÃµ½AµÄ×ø±ê£¬½øÒ»²½ÇóµÃÏòÁ¿$\overrightarrow{A{F}_{1}}¡¢\overrightarrow{A{F}_{2}}$µÄ×ø±ê£¬ÓÉ$\overrightarrow{A{F}_{1}}$•$\overrightarrow{A{F}_{2}}$=c²ÁÐʽÇóµÃÍÖÔ²µÄÀëÐÄÂÊe£®

½â´ð ½â£ºÈçͼ£¬
ÓÉ$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}$=1£¬µÃ${y}^{2}=\frac{{b}^{2}}{{a}^{2}}£¨{a}^{2}-{x}^{2}£©$£¬
È¡x=c£¬¿ÉµÃ£º${y}^{2}=\frac{{b}^{4}}{{a}^{2}}$£¬$y=¡À\frac{{b}^{2}}{a}$£¬
¡àA£¨-c£¬$\frac{{b}^{2}}{a}$£©£¬ÓÖF₁£¨-c£¬0£©£¬F₂£¨c£¬0£©£¬
Ôò$\overrightarrow{A{F}_{1}}=£¨0£¬-\frac{{b}^{2}}{a}£©£¬\overrightarrow{A{F}_{2}}=£¨2c£¬-\frac{{b}^{2}}{a}£©$£¬
ÓÉ$\overrightarrow{A{F}_{1}}$•$\overrightarrow{A{F}_{2}}$=c²£¬µÃ$£¨-\frac{{b}^{2}}{a}£©£¨-\frac{{b}^{2}}{a}£©={c}^{2}$£¬¼´$\frac{{b}^{4}}{{a}^{2}}={c}^{2}$£¬
¡à$\frac{£¨{a}^{2}-{c}^{2}£©^{2}}{{a}^{2}}={c}^{2}$£¬¼´e⁴-3e²+1=0£¬½âµÃ${e}^{2}=\frac{3-\sqrt{5}}{2}$£¬
¡à$e=\frac{\sqrt{5}-1}{2}$£®
¹ÊÑ¡£ºC£®

µãÆÀ ±¾Ì⿼²éÍÖÔ²µÄ¼òµ¥ÐÔÖÊ£¬¿¼²éÀûÓÃÏòÁ¿ÊýÁ¿»ýÇóÍÖÔ²µÄбÂÊ£¬ÌåÏÖÁËÊýÐνáºÏµÄ½âÌâ˼Ïë·½·¨£¬ÊÇÖеµÌ⣮