Question

17£®ÒÑÖªÍÖÔ²C£º$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}$=1£¨a£¾b£¾0£©µÄ×ó¡¢ÓÒ½¹µãΪF₁£¬F₂£¬PÊÇÍÖÔ²ÉÏÈÎÒâÒ»µã£¬ÇÒ|PF₁|+|PF₂|=2$\sqrt{2}$£¬ËüµÄ½¹¾àΪ2£®
£¨¢ñ£©ÇóÍÖÔ²CµÄ·½³Ì£»
£¨¢ò£©ÊÇ·ñ´æÔÚÕýʵÊýt£¬Ê¹Ö±Ïßx-y+t=0ÓëÍÖÔ²C½»ÓÚ²»Í¬µÄÁ½µãA£¬B£¬ÇÒÏ߶ÎABµÄÖеãÔÚÔ²x²+y²=$\frac{5}{6}$ÉÏ£¬Èô´æÔÚ£¬ÇótµÄÖµ£»Èô²»´æÔÚ£¬Çë˵Ã÷ÀíÓÉ£®

Answer 1

·ÖÎö £¨1£©Ö±½Ó¸ù¾ÝÍÖÔ²µÄ¶¨Òå¿ÉÇó³öa£¬ÔÙÀûÓÃa²=b²+c²Çó³öc¼´¿É£»
£¨2£©ÁªÁ¢·½³Ì×éÀûÓÃΤ´ï¶¨ÀíÇó³öx₁+x₂=$-\frac{4t}{3}$£¬y₁+y₂=x₁+x₂+2t=$\frac{2t}{3}$£¬´øÈëÖеã×ø±êµ½Ô²·½³Ì¼´¿ÉÇó³ötÖµ£®

½â´ð ½â£º£¨1£©ÒòΪF₁£¬F₂Ϊ×ó¡¢ÓÒ½¹µã£¬PÊÇÍÖÔ²ÉÏÈÎÒâÒ»µã£¬ÇÒ|PF₁|+|PF₂|=2$\sqrt{2}$
¡àa=$\sqrt{2}$£»
¡ß2c=2⇒c=1£»
¡àb=$\sqrt{{a}^{2}-{c}^{2}}$=1£»
ËùÒÔ£¬ÍÖÔ²·½³ÌΪ£º$\frac{{x}^{2}}{2}+{y}^{2}=1$
£¨2£©ÉèA£¨x₁£¬y₁£©£¬B£¨x₂£¬y₂£©
ÁªÁ¢·½³Ì×é$\left\{\begin{array}{l}{x-y+t=0}\\{\frac{{x}^{2}}{2}+{y}^{2}=1}\end{array}\right.$£¬»¯¼òºóÓУº3x²+4tx+2t²-2=0  ¢Ù£»
ÓÉ¢ÙÖª£ºx₁+x₂=$-\frac{4t}{3}$ 
ËùÒÔ£ºy₁+y₂=x₁+x₂+2t=$\frac{2t}{3}$£»
ÓÉÓÚÏ߶ÎABµÄÖеãÔÚÔ²x²+y²=$\frac{5}{6}$ÉÏ£¬
ËùÒÔÓУº$£¨-\frac{2t}{3}£©^{2}+£¨\frac{t}{3}£©^{2}=\frac{5}{6}$⇒t=¡À$\frac{\sqrt{6}}{2}$£¨¸ºÉᣩ£»
¹Ê´æÔÚt=$\frac{\sqrt{6}}{2}$Âú×ãÌâÒâ

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁËÍÖÔ²·½³Ì»ù´¡¶¨Ò壬ÒÔ¼°Î¤´ï¶¨ÀíÔÚÍÖÔ²ÓëÖ±Ïß×ÛºÏÖеÄÓ¦Óã¬ÊôÖеÈÌ⣮