Question

7£®ÔÚÖ±½Ç×ø±êϵxOyÖУ¬ÇúÏßCµÄ²ÎÊý·½³ÌΪ$\left\{\begin{array}{l}{x=\sqrt{3}cos¦Á}\\{y=sin¦Á}\end{array}\right.$£¨¦ÁΪ²ÎÊý£©£®ÒÔµãOΪ¼«µã£¬xÖáÕý°ëÖáΪ¼«ÖὨÁ¢¼«×ø±êϵ£¬Ö±ÏßlµÄ¼«×ø±ê·½³ÌΪ¦Ñcos£¨¦È-$\frac{¦Ð}{4}$£©=2$\sqrt{2}$
£¨¢ñ£©½«Ö±Ïßl»¯ÎªÖ±½Ç×ø±ê·½³Ì£»
£¨¢ò£©ÇóÇúÏßCÉϵÄÒ»µãQ µ½Ö±Ïßl µÄ¾àÀëµÄ×î´óÖµ¼°´ËʱµãQµÄ×ø±ê£®

Answer 1

·ÖÎö £¨¢ñ£©Ö±ÏßlµÄ¼«×ø±ê·½³Ìת»¯Îª¦Ñcos¦È+¦Ñsin¦È=4£¬ÓÉx=¦Ñcos¦È£¬y=¦Ñsin¦È£¬ÄÜʾ³öÖ±ÏßlµÄÖ±½Ç×ø±ê·½³Ì£®
£¨¢ò£©ÉèµãQµÄ×ø±êΪ£¨$\sqrt{3}cos¦Á£¬sin¦Á$£©£¬µãQµ½Ö±ÏßlµÄ¾àÀëΪd=$\frac{|2sin£¨¦Á+\frac{¦Ð}{3}£©-4|}{\sqrt{2}}$£¬ÓÉ´ËÄÜÇó³öÇúÏßCÉϵÄÒ»µãQ µ½Ö±Ïßl µÄ¾àÀëµÄ×î´óÖµ¼°´ËʱµãQµÄ×ø±ê£®

½â´ð ½â£º£¨¢ñ£©¡ßÖ±ÏßlµÄ¼«×ø±ê·½³ÌΪ¦Ñcos£¨¦È-$\frac{¦Ð}{4}$£©=2$\sqrt{2}$
¡à¦Ñ£¨cos$¦Ècos\frac{¦Ð}{4}$+sin$¦Èsin\frac{¦Ð}{4}$£©=2$\sqrt{2}$£¬
»¯¼òµÃ£¬¦Ñcos¦È+¦Ñsin¦È=4£¬¡­£¨1·Ö£©
ÓÉx=¦Ñcos¦È£¬y=¦Ñsin¦È£¬
¡àÖ±ÏßlµÄÖ±½Ç×ø±ê·½³ÌΪx+y=4£®¡­£¨3·Ö£©
£¨¢ò£©ÓÉÓÚµãQÊÇÇúÏßCÉϵĵ㣬Ôò¿ÉÉèµãQµÄ×ø±êΪ£¨$\sqrt{3}cos¦Á£¬sin¦Á$£©£¬¡­£¨4·Ö£©
µãQµ½Ö±ÏßlµÄ¾àÀëΪd=$\frac{|\sqrt{3}cos¦Á+sin¦Á-4|}{\sqrt{2}}$ ¡­£¨5·Ö£©
=$\frac{|2sin£¨¦Á+\frac{¦Ð}{3}£©-4|}{\sqrt{2}}$£®¡­£¨7·Ö£©
µ±sin£¨$¦Á+\frac{¦Ð}{3}$£©=-1ʱ£¬¼´$¦Á=2k¦Ð-\frac{5}{6}¦Ð$£¬
d_max=$\frac{6}{\sqrt{2}}$=3$\sqrt{2}$£®¡­£¨9·Ö£©
´Ëʱ£¬cos$¦Á=cos£¨-\frac{5}{6}¦Ð£©$=-$\frac{\sqrt{3}}{2}$£¬sin$¦Á=sin£¨-\frac{5}{6}¦Ð£©=-\frac{1}{2}$£¬
¡àµãQ£¨-$\frac{3}{2}£¬-\frac{1}{2}$£©£®¡­£¨10·Ö£©

µãÆÀ ±¾Ì⿼²éÖ±ÏßµÄÖ±½Ç×ø±ê·½³ÌµÄÇ󷨣¬¿¼²éÇúÏßÉϵÄÒ»µãµ½Ö±ÏߵľàÀëµÄ×î´óÖµ¼°´ËʱµãµÄ×ø±êµÄÇ󷨣¬ÊÇÖеµÌ⣬½âÌâʱҪÈÏÕæÉóÌ⣬עÒ⼫×ø±êÓëÖ±½Ç×ø±êµÄ»¥»¯¹«Ê½µÄºÏÀíÔËÓã®