Question

20£®ÔÚƽÃæÖ±½Ç×ø±êϵxOyÖУ¬Ö±ÏßlµÄ²ÎÊý·½³ÌΪ$\left\{\begin{array}{l}{x=-1+\frac{\sqrt{2}}{2}t}\\{y=2+\frac{\sqrt{2}}{2}t}\end{array}\right.$ £¨tΪ²ÎÊý£©£¬ÇúÏßCµÄ²ÎÊý·½³ÌΪ$\left\{\begin{array}{l}{x=4cos¦È}\\{y=cos2¦È}\end{array}\right.$£¨¦ÈΪ²ÎÊý£©£®
£¨1£©½«ÇúÏßCµÄ²ÎÊý·½³Ì»¯ÎªÆÕͨ·½³Ì£»
£¨2£©ÇóÇúÏßCÉϵĵ㵽ֱÏßlµÄ¾àÀëµÄ×î´óÖµ£®

Answer 1

·ÖÎö £¨1£©ÓÉÇúÏßCµÄ²ÎÊý·½³ÌΪ$\left\{\begin{array}{l}{x=4cos¦È}\\{y=cos2¦È}\end{array}\right.$£¨¦ÈΪ²ÎÊý£©£¬ÀûÓñ¶½Ç¹«Ê½¿ÉµÃy=cos2¦È=2cos²¦È-1=$2¡Á£¨\frac{x}{4}£©^{2}$-1£¬»¯¼òÕûÀí¿ÉµÃÇúÏßCµÄÆÕͨ·½³Ì£¬×¢ÒâxµÄÈ¡Öµ·¶Î§£®
£¨2£©Ö±ÏßlµÄÆÕͨ·½³ÌΪx-y+3=0£¬ÀûÓõ㵽ֱÏߵľàÀ빫ʽ¿ÉµÃ£ºÇúÏßCÉϵĵ㵽lµÄ¾àÀëd=$\frac{|4cos¦È-cos2¦È+3|}{\sqrt{2}}$=$\frac{|2£¨cos¦È-1£©^{2}-6|}{\sqrt{2}}$£¬¼´¿ÉµÃ³ö£®

½â´ð ½â£º£¨1£©¡ßÇúÏßCµÄ²ÎÊý·½³ÌΪ$\left\{\begin{array}{l}{x=4cos¦È}\\{y=cos2¦È}\end{array}\right.$£¨¦ÈΪ²ÎÊý£©£¬
¡ày=cos2¦È=2cos²¦È-1=$2¡Á£¨\frac{x}{4}£©^{2}$-1£¬
»¯Îªy=$\frac{{x}^{2}}{8}$-1£¬cos¦È¡Ê[-1£¬1]£¬¿ÉµÃx¡Ê[-1£¬1]£®
¡àÇúÏßCµÄÆÕͨ·½³ÌΪ£ºy=$\frac{{x}^{2}}{8}$-1£¬x¡Ê[-1£¬1]£®
£¨2£©Ö±ÏßlµÄÆÕͨ·½³ÌΪx-y+3=0£¬ÇúÏßCÉϵĵ㵽lµÄ¾àÀëd=$\frac{|4cos¦È-cos2¦È+3|}{\sqrt{2}}$=$\frac{|2£¨cos¦È-1£©^{2}-6|}{\sqrt{2}}$¡Ü$\frac{6}{\sqrt{2}}$=3$\sqrt{2}$£¬
µ±cos¦È=1ʱ£¬dÈ¡µÃ×î´óÖµ3$\sqrt{2}$£®

µãÆÀ ±¾Ì⿼²éÁ˲ÎÊý·½³Ì»¯ÎªÆÕͨ·½³Ì¡¢±¶½Ç¹«Ê½¡¢ºÍ²î¹«Ê½¡¢µãµ½Ö±ÏߵľàÀ빫ʽ£¬¿¼²éÁËÍÆÀíÄÜÁ¦Óë¼ÆËãÄÜÁ¦£¬ÊôÓÚÖеµÌ⣮