Question

16£®ÒÑÖªÍÖÔ²C£º$\frac{{x}^{2}}{{a}^{2}}$+$\frac{{y}^{2}}{{b}^{2}}$=1£¨a£¾b£¾0£©µÄÀëÐÄÂÊΪ$\frac{\sqrt{2}}{2}$£¬A£¬B·Ö±ðΪ×ó¡¢ÓÒ¶¥µã£¬F₂ΪÆäÓÒ½¹µã£¬PÊÇÍÖÔ²CÉÏÒìÓÚA£¬BµÄ¶¯µã£¬ÇÒ$\overrightarrow{PA}$•$\overrightarrow{PB}$µÄ×îСֵΪ-2£®
£¨¢ñ£©ÇóÍÖÔ²CµÄ·½³Ì£»
£¨¢ò£©Èô¹ý×ó½¹µãF₁µÄÖ±Ïß½»ÍÖÔ²ÓÚM£¬NÁ½µã£¬Çó$\overrightarrow{{F}_{2}M}$•$\overrightarrow{{F}_{2}N}$µÄÈ¡Öµ·¶Î§£®

Answer 1

·ÖÎö £¨1£©ÓÉÍÖÔ²µÄÀëÐÄÂʵõ½a£¬bµÄ¹Øϵ£¬ÔÙÓÉ$\overrightarrow{PA}$•$\overrightarrow{PB}$µÄ×îСֵΪ-2ÇóµÃaµÄÖµ£¬Ôòb¿ÉÇó£¬ÍÖÔ²·½³Ì¿ÉÇó£»
£¨2£©ÓÉ£¨1£©ÖªF₁£¨-$\sqrt{2}$£¬0£©£¬F₂£¨$\sqrt{2}$£¬0£©£¬ÔòбÂʲ»´æÔÚʱ£¬ÓÃ×ø±ê·Ö±ð±íʾ³ö$\overrightarrow{{F}_{2}M}$£¬$\overrightarrow{{F}_{2}N}$µÄ£¬Ö±½ÓÇóµÃ$\overrightarrow{{F}_{2}M}$•$\overrightarrow{{F}_{2}N}$£»Ö±ÏßбÂÊ´æÔÚʱ£¬ÉèÖ±ÏßMNµÄ·½³ÌΪy=k£¨x+$\sqrt{2}$£©£¬´úÈëÍÖÔ²·½³Ì$\frac{{x}^{2}}{4}+\frac{{y}^{2}}{2}=1$£¬ÏûÈ¥yµÃ£¨1+2k²£©x²+4$\sqrt{2}$k²x+4£¨k²-1£©=0£®ÀûÓøùÓëϵÊýµÄ¹ØϵÇóµÃM£¬NµÄºá×Ý×ø±êµÄ»ý£¬°Ñ$\overrightarrow{{F}_{2}M}$•$\overrightarrow{{F}_{2}N}$ת»¯ÎªM£¬NµÄºá×ø±êµÄºÍÓë»ýµÄÐÎʽ£¬´úÈëºó»¯Îª¹ØÓÚkµÄº¯ÊýʽµÃ´ð°¸£®

½â´ð ½â£º£¨1£©ÓÉÌâÒâÖª£¬$\frac{c}{a}=\frac{\sqrt{2}}{2}$£¬¼´$\frac{{c}^{2}}{{a}^{2}}=\frac{1}{2}$£¬
¡à$\frac{{a}^{2}-{b}^{2}}{{a}^{2}}=\frac{1}{2}$£¬Ôòa²=2b²£¬
ÉèP£¨x£¬y£©£¬
¡ß$\overrightarrow{PA}$•$\overrightarrow{PB}$=£¨-a-x£¬-y£©•£¨a-x£¬-y£©=x²-a²+y²=${x}^{2}-{a}^{2}+\frac{{a}^{2}}{2}-\frac{{x}^{2}}{2}=\frac{1}{2}£¨{x}^{2}-{a}^{2}£©$£¬
¡ß-a¡Üx¡Üa£¬¡àµ±x=0ʱ£¬$£¨\overrightarrow{PA}•\overrightarrow{PB}£©_{min}=-\frac{{a}^{2}}{2}=-2$£¬
¡àa²=4£¬Ôòb²=2£®
¡àÍÖÔ²CµÄ·½³ÌΪ$\frac{{x}^{2}}{4}+\frac{{y}^{2}}{2}=1$£»
£¨2£©ÓÉa²=4£¬b²=2£¬µÃc=$\sqrt{{a}^{2}-{b}^{2}}=\sqrt{2}$£¬
¡à${F}_{1}£¨-\sqrt{2}£¬0£©£¬{F}_{2}£¨\sqrt{2}£¬0£©$£®
ÔòÖ±ÏßбÂʲ»´æÔÚʱ£¬
M£¨-$\sqrt{2}$£¬1£©£¬N£¨-$\sqrt{2}$£¬-1£©£¬ÓÚÊÇ$\overrightarrow{{F}_{2}M}$=£¨-2$\sqrt{2}$£¬1£©£¬$\overrightarrow{{F}_{2}N}$=£¨-2$\sqrt{2}$£¬-1£©£¬
¡à$\overrightarrow{{F}_{2}M}$•$\overrightarrow{{F}_{2}N}$=7£»
Ö±ÏßбÂÊ´æÔÚʱ£¬ÉèÖ±ÏßMNµÄ·½³ÌΪy=k£¨x+$\sqrt{2}$£©£¬´úÈëÍÖÔ²·½³Ì$\frac{{x}^{2}}{4}+\frac{{y}^{2}}{2}=1$£¬ÏûÈ¥yµÃ
£¨1+2k²£©x²+4$\sqrt{2}$k²x+4£¨k²-1£©=0£®
ÉèM£¨x₁£¬y₁£©£¬N£¨x₂£¬y₂£©£¬
Ôò${x}_{1}+{x}_{2}=-\frac{4\sqrt{2}{k}^{2}}{1+2{k}^{2}}£¬{x}_{1}{x}_{2}=\frac{4£¨{k}^{2}-1£©}{1+2{k}^{2}}$£¬
¡ß$\overrightarrow{{F}_{2}M}=£¨{x}_{1}-\sqrt{2}£¬{y}_{1}£©£¬\overrightarrow{{F}_{2}N}=£¨{x}_{2}-\sqrt{2}£¬{y}_{2}£©$£¬
¡à$\overrightarrow{{F}_{2}M}$•$\overrightarrow{{F}_{2}N}$=${x}_{1}{x}_{2}-\sqrt{2}£¨{x}_{1}+{x}_{2}£©+2+{k}^{2}£¨{x}_{1}+\sqrt{2}£©£¨{x}_{2}+\sqrt{2}£©$
=$£¨1+{k}^{2}£©{x}_{1}{x}_{2}+£¨\sqrt{2}{k}^{2}-\sqrt{2}£©£¨{x}_{1}+{x}_{2}£©$+2k²+2
=$£¨1+{k}^{2}£©\frac{4£¨{k}^{2}-1£©}{1+2{k}^{2}}+\sqrt{2}£¨{k}^{2}-1£©•\frac{-4\sqrt{2}{k}^{2}}{1+2{k}^{2}}+2{k}^{2}+2$
=7-$\frac{9}{1+2{k}^{2}}$£®
¡ß1+2k²¡Ý1£¬¡à0£¼$\frac{1}{1+2{k}^{2}}$¡Ü1£¬
¡à7-$\frac{9}{1+2{k}^{2}}$¡Ê[-2£¬7£©£¬
×ÛÉÏÖª£¬$\overrightarrow{{F}_{2}M}$•$\overrightarrow{{F}_{2}N}$¡Ê[-2£¬7]£®

µãÆÀ ±¾ÌâÒÔÏòÁ¿ÎªÔØÌ壬¿¼²éÍÖÔ²µÄ±ê×¼·½³Ì£¬¿¼²éÏòÁ¿µÄÊýÁ¿»ý£¬¿¼²éÔËËãÄÜÁ¦£¬½âÌâʱӦעÒâ·ÖÀàÌÖÂÛ£¬Í¬Ê±ÕýÈ·ÓÃ×ø±ê±íʾÏòÁ¿£¬ÊÇÖеµÌâ