ÌâÄ¿ÄÚÈÝ

17£®Èçͼ£¬ÔÚÆ½ÃæÖ±½Ç×ø±êϵÖУ¬µãOÎª×ø±êÔ­µã£¬Ö±Ïßy=kx+3ÓëxÖá½»ÓÚµãA£¬ÓëyÖá½»ÓÚµãC£¬¹ýµãCµÄÅ×ÎïÏß$y=\frac{1}{2}{x^2}+bx+c$ÓëÖ±ÏßAC½»ÓÚÁíÒ»µãB£¬µãB×ø±êΪ£¨$\frac{7}{2}$£¬$\frac{45}{8}$£©£®
£¨1£©ÇóÖ±ÏߺÍÅ×ÎïÏߵĽâÎöʽ£»
£¨2£©µãPÊÇÉäÏßCBÉϵÄÒ»¸ö¶¯µã£¬¹ýµãP×÷Ö±ÏßPQ¡ÍxÖᣬ´¹×ãΪµãQ£¬½»Å×ÎïÏßÓÚµãD£¬
¢Ùµ±PD=PCʱ£¬ÇóµãPµÄ×ø±ê£®
¢ÚÔÚxÖáÉϵãQµÄÓÒ²àÈ¡µãM£¬Ê¹MQ=$\frac{3}{2}$£¬ÔÚQPµÄÑÓ³¤ÏßÉÏÈ¡µãN£¬Á¬½ÓPM£¬AN£¬ÒÑÖªtan¡ÏNAQ-tan¡ÏMPQ=$\frac{3}{4}$£¬ÇóÏß¶ÎPNµÄ³¤£®

·ÖÎö £¨1£©ÏÈÀûÓÃy=kx+3È·¶¨Cµã×ø±ê£¬È»ºó°ÑCµãºÍBµã×ø±ê´úÈëy=$\frac{1}{2}$x2+bx+cµÃ¹ØÓÚb¡¢cµÄ·½³Ì×飬Ȼºó½â·½³Ì×éÇó³öb¡¢c¼´¿ÉµÃµ½Å×ÎïÏß½âÎöʽ£»
£¨2£©¢ÙÏȰÑBµã×ø±ê´úÈëy=kx+3Çó³ökµÃµ½Ö±ÏßABµÄ½âÎöʽΪy=$\frac{3}{4}$x+3£¬Èçͼ1£¬ÀûÓÃÒ»´Îº¯ÊýͼÏóºÍ¶þ´Îº¯ÊýͼÏóÉϵãµÄ×ø±êÌØÕ÷¿ÉÉèP£¨t£¬$\frac{3}{4}$t+3£©£¬ÔòD£¨t£¬$\frac{1}{2}$t2-t+3£©£¬ÔÙÓÃt·Ö±ð±íʾ³öPDºÍPC£¬ÔòÀûÓÃPD=PC¿ÉµÃµ½¹ØÓÚtµÄ·½³Ì£¬È»ºóµÃµ½¹ØÓÚtµÄÁ½¸öÒ»Ôª¶þ´Î·½³Ì£¬Ôٽⷽ³ÌÇó³öÂú×ãÌõ¼þµÄtµÄÖµ£¬´Ó¶øµÃµ½Pµã×ø±ê£»
¢ÚÈçͼ2£¬ÏÈÀûÓÃÖ±ÏßABµÄ½âÎöʽȷ¶¨Aµã×ø±ê£¬ÉèP£¨t£¬$\frac{3}{4}$t+3£©£¬Q£¨t£¬0£©£¬Ôò¿ÉÓÃt±íʾPQºÍAQ£¬ÔÙÀûÓÃÈý½Çº¯ÊýµÄ¶¨ÒåµÃ¹ØÓÚtµÄ·½³Ì£¬È»ºó½â·½³Ì¿ÉÇó³öPNµÄ³¤£®

½â´ð ½â£º£¨1£©µ±x=0ʱ£¬y=kx+3=3£¬ÔòC£¨0£¬3£©£¬
°ÑC£¨0£¬3£©£¬B£¨$\frac{7}{2}$£¬$\frac{45}{8}$£©´úÈëy=$\frac{1}{2}$x2+bx+cµÃ$\left\{\begin{array}{l}{c=3}\\{\frac{1}{2}¡Á\frac{49}{4}+\frac{7}{2}b+c=\frac{45}{8}}\end{array}\right.$£¬½âµÃ$\left\{\begin{array}{l}{b=-1}\\{c=3}\end{array}\right.$
ËùÒÔÅ×ÎïÏß½âÎöʽΪy=$\frac{1}{2}$x2-x+3£»
£¨2£©¢Ù°ÑB£¨$\frac{7}{2}$£¬$\frac{45}{8}$£©´úÈëy=kx+3µÃ$\frac{7}{2}$k+3=$\frac{45}{8}$£¬½âµÃk=$\frac{3}{4}$£¬
ËùÒÔÖ±ÏßABµÄ½âÎöʽΪy=$\frac{3}{4}$x+3£¬
Èçͼ1£¬ÉèP£¨t£¬$\frac{3}{4}$t+3£©£¬ÔòD£¨t£¬$\frac{1}{2}$t2-t+3£©£¬
ËùÒÔPD=|$\frac{1}{2}$t2-t+3-£¨$\frac{3}{4}$t+3£©|=|$\frac{1}{2}$t2-$\frac{7}{4}$t|£¬
¶øPC=$\sqrt{{t}^{2}+£¨\frac{3}{4}t+3-3£©^{2}}$=$\frac{5}{4}$t£¬
ÒòΪPD=PC£¬
ËùÒÔ|$\frac{1}{2}$t2-$\frac{7}{4}$t|=$\frac{5}{4}$t£¬
µ±$\frac{1}{2}$t2-$\frac{7}{4}$t=$\frac{5}{4}$tʱ£¬½âµÃt1=0£¨ÉáÈ¥£©£¬t2=6£¬´ËʱPµã×ø±êΪ£¨6£¬$\frac{15}{2}$£©£»
µ±$\frac{1}{2}$t2-$\frac{7}{4}$t=-$\frac{5}{4}$tʱ£¬½âµÃt1=0£¨ÉáÈ¥£©£¬t2=1£¬´ËʱPµã×ø±êΪ£¨1£¬$\frac{15}{4}$£©£»
×ÛÉÏËùÊö£¬Âú×ãÌõ¼þµÄPµã×ø±êΪ£¨6£¬$\frac{15}{2}$£©»ò£¨1£¬$\frac{15}{4}$£©£»
¢ÚÈçͼ2£¬µ±y=0ʱ£¬$\frac{3}{4}$x+3=0£¬½âµÃx=-4£¬ÔòA£¨-4£¬0£©£¬
ÉèP£¨t£¬$\frac{3}{4}$t+3£©£¬Q£¨t£¬0£©£¬ÔòPQ=$\frac{3}{4}$t+3£¬AQ=t+4£¬
ÔÚRt¡÷NAQÖУ¬tan¡ÏNAQ=$\frac{NQ}{AQ}$=$\frac{NP+\frac{3}{4}t+3}{t+4}$£¬
ÔÚRt¡÷NMQÖУ¬tan¡ÏMPQ=$\frac{QM}{PQ}$=$\frac{\frac{3}{2}}{\frac{3}{4}t+3}$£¬
¶øtan¡ÏNAQ-tan¡ÏMPQ=$\frac{3}{4}$£¬
ËùÒÔ$\frac{NP+\frac{3}{4}t+3}{t+4}$=$\frac{\frac{3}{2}}{\frac{3}{4}t+3}$£¬
ËùÒÔPN=2£®

µãÆÀ ±¾Ì⿼²éÁ˶þ´Îº¯ÊýµÄ×ÛºÏÌ⣺ÊìÁ·ÕÆÎÕ¶þ´Îº¯ÊýͼÏóÉϵãµÄ×ø±êÌØÕ÷¡¢¶þ´Îº¯ÊýµÄÐÔÖʺÍÈñ½ÇÈý½Çº¯ÊýµÄ¶¨Ò壻»áÀûÓôý¶¨ÏµÊý·¨Çóº¯Êý½âÎöʽ£»Àí½â×ø±êÓëͼÐÎÐÔÖÊ£¬¼ÇסÁ½µã¼äµÄ¾àÀ빫ʽ£»»á½âÒ»Ôª¶þ´Î·½³Ì£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø