Question

11£®ÒÑÖªÍÖÔ²C£º$\frac{{x}^{2}}{{a}^{2}}$+$\frac{{y}^{2}}{{b}^{2}}$=1£¨a£¾b£¾0£©µÄ×ó¡¢ÓÒ½¹µã·Ö±ðÊÇF₁£¨-c£¬0£©£¬F₂£¨c£¬0£©£¬µãMÊÇÍÖÔ²ÉÏÈÎÒâÒ»µã£¬¡÷MF₁F₂µÄÖܳ¤ÊÇ2$\sqrt{2}$+2£¬ÇÒ¡÷MF₁F₂Ãæ»ýµÄ×î´óÖµÊÇ1£®
£¨1£©ÇóÍÖÔ²CµÄ±ê×¼·½³Ì£»
£¨2£©ÈôNÊÇÍÖÔ²ÉÏÒ»µã£¬µãM£¬N²»Öغϣ¬OΪ×ø±êÔ­µã£¬µ±Ö±ÏßMNµÄбÂÊΪ2ʱ£¬Çó¡÷OMNÃæ»ýµÄ×î´óÖµ£®

Answer 1

·ÖÎö £¨1£©ÓɵãMÊÇÍÖÔ²ÉÏÈÎÒâÒ»µã£¬¡÷MF₁F₂µÄÖܳ¤ÊÇ2$\sqrt{2}$+2£¬ÇÒ¡÷MF₁F₂Ãæ»ýµÄ×î´óÖµÊÇ1£®¿ÉµÃ$\left\{\begin{array}{l}{2a+2c=2\sqrt{2}+2}\\{\frac{1}{2}¡Á2c¡Áb=1}\\{{a}^{2}={b}^{2}+{c}^{2}}\end{array}\right.$£¬½â³ö¼´¿ÉµÃ³ö£®
£¨2£©ÉèÖ±ÏßMNµÄ·½³ÌΪ£ºy=2x+m£¬M£¨x₁£¬y₁£©£¬N£¨x₂£¬y₂£©£®ÓëÍÖÔ²·½³ÌÁªÁ¢¿ÉµÃ£º9x²+8mx+2m²-2=0£¬¡÷£¾0£¬ÀûÓøùÓëϵÊýµÄ¹Øϵ¿ÉµÃ|MN|=$\sqrt{£¨1+{2}^{2}£©[£¨{x}_{1}+{x}_{2}£©^{2}-4{x}_{1}{x}_{2}]}$£®Ô­µãµ½Ö±ÏßMNµÄ¾àÀëd=$\frac{|m|}{\sqrt{5}}$£¬¿ÉµÃS_¡÷OMN=$\frac{1}{2}|MN|$d£¬ÔÙÀûÓûù±¾²»µÈʽµÄÐÔÖʼ´¿ÉµÃ³ö£®

½â´ð ½â£º£¨1£©¡ßµãMÊÇÍÖÔ²ÉÏÈÎÒâÒ»µã£¬¡÷MF₁F₂µÄÖܳ¤ÊÇ2$\sqrt{2}$+2£¬ÇÒ¡÷MF₁F₂Ãæ»ýµÄ×î´óÖµÊÇ1£®
¡à$\left\{\begin{array}{l}{2a+2c=2\sqrt{2}+2}\\{\frac{1}{2}¡Á2c¡Áb=1}\\{{a}^{2}={b}^{2}+{c}^{2}}\end{array}\right.$£¬½âµÃa=$\sqrt{2}$£¬b=c=1£®
¡àÍÖÔ²CµÄ±ê×¼·½³ÌΪ$\frac{{x}^{2}}{2}+{y}^{2}$=1£®
£¨2£©ÉèÖ±ÏßMNµÄ·½³ÌΪ£ºy=2x+m£¬M£¨x₁£¬y₁£©£¬N£¨x₂£¬y₂£©£®
ÁªÁ¢$\left\{\begin{array}{l}{y=2x+m}\\{{x}^{2}+2{y}^{2}=2}\end{array}\right.$£¬»¯Îª£º9x²+8mx+2m²-2=0£¬
¡÷£¾0£¬¿ÉµÃ£ºm²£¼9£®
¡àx₁+x₂=$\frac{-8m}{9}$£¬x₁x₂=$\frac{2{m}^{2}-2}{9}$£¬
¡à|MN|=$\sqrt{£¨1+{2}^{2}£©[£¨{x}_{1}+{x}_{2}£©^{2}-4{x}_{1}{x}_{2}]}$=$\sqrt{5[\frac{64{m}^{2}}{81}-\frac{4£¨2{m}^{2}-2£©}{9}]}$=$\frac{2\sqrt{5£¨18-2{m}^{2}£©}}{9}$£®
Ô­µãµ½Ö±ÏßMNµÄ¾àÀëd=$\frac{|m|}{\sqrt{5}}$£¬
¡àS_¡÷OMN=$\frac{1}{2}|MN|$d=$\frac{1}{2}$¡Á$\frac{2\sqrt{5£¨18-2{m}^{2}£©}}{9}$¡Á$\frac{|m|}{\sqrt{5}}$=$\frac{\sqrt{{m}^{2}£¨18-2{m}^{2}£©}}{9}$$¡Ü\frac{\sqrt{\frac{1}{2}£¨\frac{2{m}^{2}+18-2{m}^{2}}{2}£©^{2}}}{9}$=$\frac{\sqrt{2}}{2}$£¬µ±ÇÒ½öµ±m=¡À$\frac{3\sqrt{2}}{2}$ʱȡµÈºÅ£®
¡à¡÷OMNÃæ»ýµÄ×î´óֵΪ$\frac{\sqrt{2}}{2}$£®

µãÆÀ ±¾Ì⿼²éÁËÍÖÔ²µÄ±ê×¼·½³Ì¼°ÆäÐÔÖÊ¡¢Ö±ÏßÓëÍÖÔ²ÏཻÏÒ³¤ÎÊÌâ¡¢µãµ½Ö±ÏߵľàÀ빫ʽ¡¢Èý½ÇÐÎÃæ»ý¼ÆË㹫ʽ£¬¿¼²éÁËÍÆÀíÄÜÁ¦Óë¼ÆËãÄÜÁ¦£¬ÊôÓÚÄÑÌ⣮