Question

9£®ÒÑÖªÍÖÔ²E£º$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$£¨a£¾b£¾0£©µÄ×ó¡¢ÓÒÁ½½¹µã·Ö±ðΪF₁£¬F₂£¬¶ÌÖáµÄÒ»¸ö¶ËµãΪM£¬Ö±Ïßl£º3x-4y=0½»ÍÖÔ²EÓÚA£¬BÁ½µã£¬ÇÒ|AF₂|+|BF₂|=2$\sqrt{2}$£®
£¨1£©ÈôÍÖÔ²µÄÀëÐÄÂÊΪ$\frac{\sqrt{2}}{2}$£¬ÇóÍÖÔ²µÄ·½³Ì£»
£¨2£©ÈôµãMµ½Ö±ÏßlµÄ¾àÀ벻СÓÚ$\frac{4}{5}$£¬ÇóÍÖÔ²µÄÀëÐÄÂʵÄÈ¡Öµ·¶Î§£®

Answer 1

·ÖÎö £¨1£©Á¬½ÓAF₁£¬BF₁£¬¿ÉµÃËıßÐÎAF₂BF₁ΪƽÐÐËıßÐΣ¬ÓÉÍÖÔ²µÄ¶¨Òå¿ÉµÃ£¬2a=2$\sqrt{2}$£¬ÔÙÓÉÀëÐÄÂʹ«Ê½¿ÉµÃc£¬b£¬½ø¶øµÃµ½ÍÖÔ²µÄ·½³Ì£»
£¨2£©Éè³öM£¨0£¬b£©£¬ÔËÓõ㵽ֱÏߵľàÀ빫ʽ¿ÉµÃbµÄ·¶Î§£¬ÔÙÓÉÀëÐÄÂʹ«Ê½£¬¼´¿ÉµÃµ½ËùÇó·¶Î§£®

½â´ð ½â£º£¨1£©Á¬½ÓAF₁£¬BF₁£¬
¿ÉµÃËıßÐÎAF₂BF₁ΪƽÐÐËıßÐΣ¬
¼´ÓÐ|AF₂|+|BF₂|=|AF₂|+|AF₁|=2$\sqrt{2}$£¬
ÓÉÍÖÔ²µÄ¶¨Òå¿ÉµÃ£¬2a=2$\sqrt{2}$£¬¼´a=$\sqrt{2}$£¬
ÓÖe=$\frac{c}{a}$=$\frac{\sqrt{2}}{2}$£¬¿ÉµÃc=1£¬b=$\sqrt{{a}^{2}-{c}^{2}}$=1£®
ÔòÍÖÔ²µÄ·½³ÌΪ$\frac{{x}^{2}}{2}$+y²=1£»
£¨2£©ÓÉÌâÒâ¿ÉÉèM£¨0£¬b£©£¬
ÓɵãMµ½Ö±Ïßl£º3x-4y=0µÄ¾àÀ벻СÓÚ$\frac{4}{5}$£¬
¿ÉµÃd=$\frac{|0-4b|}{\sqrt{{3}^{2}+{4}^{2}}}$¡Ý$\frac{4}{5}$£¬¼´Îªb¡Ý1£¬
ÓÉe=$\frac{c}{a}$=$\frac{\sqrt{{a}^{2}-{b}^{2}}}{a}$=$\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}$¡Ü$\sqrt{1-\frac{1}{2}}$=$\frac{\sqrt{2}}{2}$£¬
ÔòÓÐÍÖÔ²µÄÀëÐÄÂʵķ¶Î§ÊÇ£¨0£¬$\frac{\sqrt{2}}{2}$]£®

µãÆÀ ±¾Ì⿼²éÍÖÔ²µÄ¶¨Òå¡¢·½³ÌºÍÐÔÖÊ£¬¿¼²éÀëÐÄÂʵÄÔËÓã¬Í¬Ê±¿¼²éÔËËãÄÜÁ¦£¬ÊôÓÚÖеµÌ⣮