Question

5£®¶ÔÓÚƽÃæÖ±½Ç×ø±êϵxOyÖеĵãP£¨m£¬n£©£¬¶¨ÒåÒ»Öֱ任£º×÷µãP£¨m£¬n£©¹ØÓÚyÖá¶Ô³ÆµÄµãP¡ä£¬ÔÙ½«P¡äÏò×óƽÒÆk£¨k£¾0£©¸öµ¥Î»µÃµ½µãP_k¡ä£¬P_k¡ä½Ð×ö¶ÔµãP£¨m£¬n£©µÄk½×¡°?¡±±ä»»£®ÈôÒ»¸öº¯ÊýͼÏóÉÏËùÓе㶼½øÐÐÁËk½×¡°?¡±±ä»»ºó×é³ÉµÄͼÐγÆΪ´Ëº¯Êý½øÐÐÁËk½×¡°?¡±±ä»»ºóµÄͼÐΣ®
£¨1£©ÇóP£¨3£¬2£©µÄ3½×¡°?¡±±ä»»ºóP₃¡äµÄ×ø±ê£»
£¨2£©ÈôÖ±Ïßy=x+1¾­¹ýk½×¡°?¡±±ä»»ºóµÄͼÏóÓë·´±ÈÀýº¯ÊýµÄͼÏóy=$\frac{2}{x}$ûÓй«¹²µã£¬ÇókµÄÈ¡Öµ·¶Î§£®
£¨3£©ÈôÅ×ÎïÏßC₁£ºy=x²-4x+3ÓëÖ±Ïßl£ºy=-x+3½»ÓÚA£¬BÁ½µã£¬Å×ÎïÏßC₁¾­¹ýk½×¡°?¡±±ä»»ºóµÄͼÏó¼ÇΪC₂£¬C₂ÓëÖ±Ïßl½»ÓÚC£¬DÁ½µã£¬Èô$\frac{CD}{AB}$=$\frac{7}{3}$£¬ÇókµÄÖµ£®

Answer 1

·ÖÎö £¨1£©k½×¡°?¡±±ä»»µÄ¶¨Ò壬¼´¿ÉÇó³ö3½×¡°?¡±±ä»»ºóP₃¡äµÄ×ø±ê£®
£¨2£©Ö±Ïßy=x+1¾­¹ýk½×¡°?¡±±ä»»ºóµÄ½âÎöʽΪy=-x-k+1£¬ÓÉ$\left\{\begin{array}{l}{y=\frac{2}{x}}\\{y=-x-k+1}\end{array}\right.$ÏûÈ¥yµÃµ½x²+£¨k-1£©x+2=0£¬ÓÉÌâÒâ¡÷£¼0£¬½â²»µÈʽ¼´¿É£®
£¨3£©ÓÉÌâÒâC₂µÄ½âÎöʽΪy=£¨-x-k£©²-4£¨-x-k£©+3£¬ÉèC£¨x₁£¬y₁£©£¬D£¨x₂£¬y₂£©£¬ÓÉ$\left\{\begin{array}{l}{y=-x+3}\\{y=£¨-x-k£©^{2}-4£¨-x-k£©+3}\end{array}\right.$ÏûÈ¥yµÃµ½x²+£¨2k+5£©x+k²+4k=0£¬¿ÉµÃx₁+x₂=-£¨2k+5£©£¬x₁x₂=k²+4k£¬y₂-y₁=-£¨x₂-x₁£©£¬ÍƳöCD=$\sqrt{£¨{x}_{2}-{x}_{1}£©^{2}+£¨{y}_{2}-{y}_{1}£©^{2}}$=$\sqrt{2£¨{x}_{2}-{x}_{1}£©^{2}}$=$\sqrt{2[£¨{x}_{1}+{x}_{2}£©^{2}-4{x}_{1}{x}_{2}]}$=$\sqrt{2[£¨2k+5£©^{2}-4£¨{k}^{2}+4k£©]}$=$\sqrt{8k+50}$£¬Çó³öAB£¬Áгö·½³Ì¼´¿É½â¾öÎÊÌ⣮

½â´ð ½â£º£¨1£©µãP£¨3£¬2£©¹ØÓÚyÖáµÄ¶Ô³ÆµãP¡ä£¨-3£¬2£©£¬ÔÙ½«P¡äÏò×óƽÒÆ3¸öµ¥Î»µÃµ½µãP₃¡ä£¨-6£¬2£©£®

£¨2£©Ö±Ïßy=x+1¾­¹ýk½×¡°?¡±±ä»»ºóµÄ½âÎöʽΪy=-x-k+1£¬
ÓÉ$\left\{\begin{array}{l}{y=\frac{2}{x}}\\{y=-x-k+1}\end{array}\right.$ÏûÈ¥yµÃµ½x²+£¨k-1£©x+2=0£¬
¡ßÖ±Ïßy=x+1¾­¹ýk½×¡°?¡±±ä»»ºóµÄͼÏóÓë·´±ÈÀýº¯ÊýµÄͼÏóy=$\frac{2}{x}$ûÓй«¹²µã£¬
¡à¡÷£¼0£¬
¡à£¨k-1£©²-8£¼0£¬
¡à1-2$\sqrt{2}$£¼k£¼1+2$\sqrt{2}$£®

£¨3£©ÓÉÌâÒâC₂µÄ½âÎöʽΪy=£¨-x-k£©²-4£¨-x-k£©+3£¬ÉèC£¨x₁£¬y₁£©£¬D£¨x₂£¬y₂£©£¬
ÓÉ$\left\{\begin{array}{l}{y={x}^{2}-4x+3}\\{y=-x+3}\end{array}\right.$£¬½âµÃ$\left\{\begin{array}{l}{x=0}\\{y=3}\end{array}\right.$»ò$\left\{\begin{array}{l}{x=3}\\{y=0}\end{array}\right.$£¬
²»·ÁÉèA£¨0£¬3£©£¬B£¨3£¬0£©£¬ÔòAB=3$\sqrt{2}$£¬
ÓÉ$\left\{\begin{array}{l}{y=-x+3}\\{y=£¨-x-k£©^{2}-4£¨-x-k£©+3}\end{array}\right.$ÏûÈ¥yµÃµ½x²+£¨2k+5£©x+k²+4k=0£¬
¡àx₁+x₂=-£¨2k+5£©£¬x₁x₂=k²+4k£¬y₂-y₁=-£¨x₂-x₁£©£¬
¡àCD=$\sqrt{£¨{x}_{2}-{x}_{1}£©^{2}+£¨{y}_{2}-{y}_{1}£©^{2}}$=$\sqrt{2£¨{x}_{2}-{x}_{1}£©^{2}}$=$\sqrt{2[£¨{x}_{1}+{x}_{2}£©^{2}-4{x}_{1}{x}_{2}]}$=$\sqrt{2[£¨2k+5£©^{2}-4£¨{k}^{2}+4k£©]}$=$\sqrt{8k+50}$£¬
¡ß$\frac{CD}{AB}$=$\frac{7}{3}$£¬
¡à$\frac{8K+50}{18}$=$\frac{49}{9}$£¬
¡àk=6£®

µãÆÀ ±¾Ì⿼²é¶þ´Îº¯Êý×ÛºÏÌâ¡¢k½×¡°?¡±±ä»»µÄ¶¨Òå¡¢Ò»´Îº¯Êý¡¢Ò»Ôª¶þ´Î·½³Ì¸ùÓëϵÊý¹Øϵ¡¢Á½µã¼ä¾àÀ빫ʽµÈ֪ʶ£¬½âÌâµÄ¹Ø¼üÊÇÀí½âÌâÒ⣬ѧ»áÇók½×¡°?¡±±ä»»ºóµÄÖ±Ïß¡¢Å×ÎïÏߵĽâÎöʽÊÇÄѵ㣬ѧ»áÀûÓ÷½³Ì×éÇó½»µã×ø±ê£¬ÊôÓÚÖп¼Ñ¹ÖáÌ⣮