Question

13£®Èçͼ£¬ÒÑÖªÖ±Ïßy=mx+nÓë·´±ÈÀýº¯Êýy=$\frac{k}{x}$½»ÓÚA¡¢BÁ½µã£¬µãAÔÚµãBµÄ×ó±ß£¬ÓëxÖá¡¢yÖá·Ö±ð½»µãC¡¢µãD£¬AE¡ÍxÖáÓÚE£¬BF¡ÍyÖáÓÚF£®
£¨1£©Èôm=k£¬n=0£¬ÇóA¡¢B£»Á½µãµÄ×ø±ê£»
£¨2£©Èçͼ1£¬ÈôA£¨x₁£¬y₁£©¡¢B£¨x₂£¬y₂£©£¬Ð´³öy₁+y₂ÓënµÄ´óС¹Øϵ£¬²¢Ö¤Ã÷£»
£¨3£©Èçͼ2£¬M¡¢N·Ö±ðΪ·´±ÈÀýº¯Êýy=$\frac{b}{x}$ͼÏóÉϵĵ㣬AM¡ÎMN¡ÎxÖᣮÈô$\frac{1}{AM}$+$\frac{1}{BN}$=$\frac{5}{3}$£¬ÇÒAM¡¢BNÖ®¼äµÄ¾àÀëΪ5£¬Ôòk-b=3£®

Answer 1

·ÖÎö £¨1£©ÓÉ$\left\{\begin{array}{l}{y=kx}\\{y=\frac{k}{x}}\end{array}\right.$½âµÃ£º$\left\{\begin{array}{l}{x=1}\\{y=k}\end{array}\right.$ºÍ$\left\{\begin{array}{l}{x=-1}\\{y=-k}\end{array}\right.$£¬¼´¿ÉµÃ³ö´ð°¸£»
£¨2£©°ÑÈôA£¨x₁£¬y₁£©¡¢B£¨x₂£¬y₂£©´úÈëy=mx+nµÃ³öy₁+y₂=m£¨x₁+x₂£©+2n£¬ÓÉ$\left\{\begin{array}{l}{y=mx+n}\\{y=\frac{k}{x}}\end{array}\right.$µÃ³ömx²+nx-k=0£¬ÓɸùÓëϵÊý¹ØϵµÃ³öx₁+x₂=-$\frac{n}{m}$£¬µÃ³öy₁+y₂=n¼´¿É£»
£¨3£©ÉèA£¨x₁£¬y₁£©¡¢B£¨x₂£¬y₂£©£¬ÔòM£¨x₃£¬y₁£©¡¢N£¨x₄£¬y₂£©£¬ÓÉ·´±ÈÀýº¯ÊýµÃ³öx₁y₁=x₂y₂=k£¬x₃y₁=x₄y₂=b£¬ÓÉ×ø±êÓëͼÐÎÐÔÖʵóöAM=x₃-x₁=$\frac{b}{{y}_{1}}$-$\frac{k}{{y}_{1}}$=$\frac{b-k}{{y}_{1}}$£¬bn=x₂-x₄=$\frac{k}{{y}_{2}}$-$\frac{b}{{y}_{2}}$=$\frac{k-b}{{y}_{2}}$£¬y₂-y₁=5£¬ÔÙ½áºÏÒÑÖªÌõ¼þ£¬¼´¿ÉµÃ³ö´ð°¸£®

½â´ð ½â£º£¨1£©ÓÉ$\left\{\begin{array}{l}{y=kx}\\{y=\frac{k}{x}}\end{array}\right.$½âµÃ£º$\left\{\begin{array}{l}{x=1}\\{y=k}\end{array}\right.$ºÍ$\left\{\begin{array}{l}{x=-1}\\{y=-k}\end{array}\right.$£¬
¡àA£¨1£®k£©£¬B£¨-1£¬-k£©£»
£¨2£©y₁+y₂=n£»ÀíÓÉÈçÏ£º
°ÑÈôA£¨x₁£¬y₁£©¡¢B£¨x₂£¬y₂£©´úÈëy=mx+nÖУ¬
Ôòy₁=mx₁+n£¬y₂=mx₂+n£¬
¡ày₁+y₂=m£¨x₁+x₂£©+2n£¬
ÓÉ$\left\{\begin{array}{l}{y=mx+n}\\{y=\frac{k}{x}}\end{array}\right.$µÃ£ºmx+n=$\frac{k}{x}$£¬
¡àmx²+nx-k=0£¬
¡ßÖ±Ïßy=mx+nÓë·´±ÈÀýº¯Êýy=$\frac{k}{x}$½»ÓÚA¡¢BÁ½µã£¬
¡àx₁+x₂=-$\frac{n}{m}$£¬
¡ày₁+y₂=m•£¨-$\frac{n}{m}$£©+2n=-n+2n=n£»
£¨3£©ÉèA£¨x₁£¬y₁£©¡¢B£¨x₂£¬y₂£©£¬
¡ßAM¡ÎMN¡ÎxÖᣮ¡àM£¨x₃£¬y₁£©¡¢N£¨x₄£¬y₂£©£¬
¡àx₁y₁=x₂y₂=k£¬x₃y₁=x₄y₂=b£¬AM=x₃-x₁=$\frac{b}{{y}_{1}}$-$\frac{k}{{y}_{1}}$=$\frac{b-k}{{y}_{1}}$£¬bn=x₂-x₄=$\frac{k}{{y}_{2}}$-$\frac{b}{{y}_{2}}$=$\frac{k-b}{{y}_{2}}$£¬y₂-y₁=5£¬
¡ß$\frac{1}{AM}$+$\frac{1}{BN}$=$\frac{5}{3}$£¬
¡à$\frac{{y}_{1}}{b-k}$+$\frac{{y}_{2}}{k-b}$=$\frac{5}{3}$£¬
¡à$\frac{{y}_{2}-{y}_{1}}{k-b}$=$\frac{5}{3}$£¬
¡à$\frac{5}{k-b}$=$\frac{5}{3}$£¬
¡àk-b=3£»
¹Ê´ð°¸Îª£º3£®

µãÆÀ ±¾ÌâÊÇ·´±ÈÀýº¯Êý×ÛºÏÌâÄ¿£¬¿¼²éÁË·´±ÈÀýº¯ÊýµÄÓ¦Óá¢×ø±êÓëͼÐÎÐÔÖÊ¡¢·½³Ì×éµÄ½â·¨¡¢¸ùÓëϵÊý¹ØϵµÈ֪ʶ£»±¾Ìâ×ÛºÏÐÔÇ¿£¬ÓÐÒ»¶¨ÄѶȣ®