Question

8£®ÔÚÖ±½Ç×ø±êϵxOyÖУ¬ÇúÏßCµÄ²ÎÊý·½³ÌÊÇ$\left\{\begin{array}{l}{x=2cos¦Á}\\{y=sin¦Á}\end{array}\right.$£¨¦ÁΪ²ÎÊý£©£¬ÒÔ×ø±êÔ­µãΪ¼«µã£¬ÒÔxÖáµÄÕý°ëÖáΪ¼«Öᣬ½¨Á¢¼«×ø±êϵ£¬Ö±ÏßlµÄ¼«×ø±ê·½³ÌΪ¦Ñsin£¨¦È+$\frac{¦Ð}{4}$£©=$\sqrt{2}$£®
£¨¢ñ£©½«ÇúÏßCµÄ²ÎÊý·½³Ì»¯ÎªÆÕͨ·½³Ì£¬½«Ö±ÏßlµÄ¼«×ø±ê·½³Ì»¯ÎªÖ±½Ç×ø±ê·½³Ì£»
£¨¢ò£©ÉèµãPÔÚÇúÏßCÉÏ£¬ÇóµãPµ½Ö±ÏßlµÄ×î´ó¾àÀ룮

Answer 1

·ÖÎö £¨¢ñ£©ÀûÓÃÈýÖÖ·½³ÌµÄת»¯·½·¨£¬¼´¿ÉµÃ³ö½áÂÛ£»
£¨¢ò£©ÉèP£¨2cos¦È£¬sin¦È£©£¬ÀûÓõ㵽ֱÏߵľàÀ빫ʽ¼°ÕýÏÒº¯ÊýµÄµ¥µ÷ÐÔ¼´¿ÉµÃ³ö£®

½â´ð ½â£º£¨¢ñ£©ÇúÏßCµÄ²ÎÊý·½³ÌÊÇ$\left\{\begin{array}{l}{x=2cos¦Á}\\{y=sin¦Á}\end{array}\right.$£¨¦ÁΪ²ÎÊý£©£¬ÆÕͨ·½³ÌΪ$\frac{{x}^{2}}{4}+{y}^{2}$=1£»
Ö±ÏßlµÄ¼«×ø±ê·½³ÌΪ¦Ñsin£¨¦È+$\frac{¦Ð}{4}$£©=$\sqrt{2}$£¬¼´$\frac{\sqrt{2}}{2}£¨¦Ñsin¦È+¦Ñcos¦È£©=\sqrt{2}$£¬Ö±½Ç×ø±ê·½³ÌΪx+y-2=0£»
£¨¢ò£©ÉèP£¨2cos¦È£¬sin¦È£©£¬
ÔòµãPµ½Ö±Ïßl¾àÀëd=$\frac{|2cos¦È+sin¦È-2|}{\sqrt{2}}$=$\frac{|\sqrt{5}sin£¨¦È+¦Á£©-2|}{\sqrt{2}}$£®
¡àµãPµ½Ö±Ïßl¾àÀëµÄ×î´óֵΪ$\frac{2+\sqrt{5}}{\sqrt{2}}$=$\sqrt{2}$+$\frac{\sqrt{10}}{2}$£®

µãÆÀ ±¾Ì⿼²éÁ˼«×ø±ê·½³Ì»¯ÎªÖ±½Ç×ø±ê·½³Ì¡¢²ÎÊý·½³Ì»¯ÎªÆÕͨ·½³Ì¡¢µãµ½Ö±ÏߵľàÀ빫ʽ¼°ÕýÏÒº¯ÊýµÄµ¥µ÷ÐÔ£¬ÊôÓÚÖеµÌ⣮