Question

9£®ÒÑÖªÇúÏß${C_1}£º\left\{\begin{array}{l}x=1+\frac{{\sqrt{2}}}{2}t\\ y=\frac{{\sqrt{2}}}{2}t\end{array}\right.$£¨tΪ²ÎÊý£©£¬ÒÔÔ­µãΪ¼«µã£¬ÒÔxÕý°ëÖáΪ¼«Öᣬ½¨Á¢¼«×ø±êϵ£¬ÇúÏß${C_2}£º\frac{1}{¦Ñ^2}=\frac{{{{cos}^2}¦È}}{2}+{sin^2}¦È$£®
£¨¢ñ£©Ð´³öÇúÏßC₁µÄÆÕͨ·½³Ì£¬ÇúÏßC₂µÄÖ±½Ç×ø±ê·½³Ì£»
£¨¢ò£©ÈôM£¨1£¬0£©£¬ÇÒÇúÏßC₁ÓëÇúÏßC₂½»ÓÚÁ½¸ö²»Í¬µÄµãA£¬B£¬Çó$\frac{|MA|•|MB|}{|AB|}$µÄÖµ£®

Answer 1

·ÖÎö £¨¢ñ£©ÏûÈ¥²ÎÊýt£¬¼´¿ÉÇóµÃC₁µÄÆÕͨ·½³Ì£¬ÓÉ$\left\{\begin{array}{l}{x=¦Ñcos¦È}\\{y=¦Ñsin¦È}\end{array}\right.$£¬»¯¼ò¼´¿ÉÇóµÃÇúÏßC₂µÄÖ±½Ç×ø±ê·½³Ì£»
£¨¢ò£©½«ÇúÏßC₁´úÈëÇúÏßC₂µÄ·½³Ì£¬ÇóµÃAºÍBµã×ø±ê£¬¸ù¾ÝÁ½µãÖ®¼äµÄ¾àÀ빫ʽ£¬¼´¿ÉÇóµÃ$\frac{|MA|•|MB|}{|AB|}$µÄÖµ£®

½â´ð ½â£º£¨¢ñ£©½«y=$\frac{\sqrt{2}}{2}$t£¬´úÈëx=1+$\frac{\sqrt{2}}{2}$t£¬ÕûÀíµÃx-y-1=0£¬ÔòÇúÏßC₁µÄÆÕͨ·½x-y-1=0£»
ÇúÏß${C_2}£º\frac{1}{¦Ñ^2}=\frac{{{{cos}^2}¦È}}{2}+{sin^2}¦È$£¬Ôò1=$\frac{{¦Ñ}^{2}co{s}^{2}¦È}{2}$+¦Ñ²sin²¦È£®
ÓÉ$\left\{\begin{array}{l}{x=¦Ñcos¦È}\\{y=¦Ñsin¦È}\end{array}\right.$£¬ÔòÇúÏßC₂µÄÖ±½Ç×ø±ê·½³Ì$\frac{{x}^{2}}{2}+{y}^{2}=1$£»
£¨¢ò£©ÓÉ$\left\{\begin{array}{l}{y=x-1}\\{\frac{{x}^{2}}{2}+{y}^{2}=1}\end{array}\right.$£¬ÕûÀíµÃ£º3x²-4x=0£¬½âµÃ£ºx=0»òx=$\frac{4}{3}$£¬
ÔòA£¨0£¬-1£©£¬B£¨$\frac{4}{3}$£¬$\frac{1}{3}$£©£¬
¡àØ­MAØ­=$\sqrt{£¨1-0£©^{2}+£¨0+1£©^{2}}$=$\sqrt{2}$£¬Ø­MBØ­=$\sqrt{£¨\frac{4}{3}-1£©^{2}+£¨\frac{1}{3}-0£©^{2}}$=$\frac{\sqrt{2}}{3}$£¬
¡àØ­ABØ­=$\sqrt{£¨\frac{4}{3}-0£©^{2}+£¨\frac{1}{3}+1£©^{2}}$=$\frac{4\sqrt{2}}{3}$£¬
¡à$\frac{|MA|•|MB|}{|AB|}$=$\frac{\sqrt{2}¡Á\frac{\sqrt{2}}{3}}{\frac{4\sqrt{2}}{3}}$=$\frac{{\sqrt{2}}}{4}$£¬
¡à$\frac{|MA|•|MB|}{|AB|}$µÄÖµ$\frac{{\sqrt{2}}}{4}$£®

µãÆÀ ±¾Ì⿼²éÖ±ÏߵIJÎÊý·½³Ì£¬ÍÖÔ²µÄ¼«×ø±ê·½³Ì£¬Ö±ÏßÓëÍÖÔ²µÄλÖùØϵ£¬¿¼²é¼ÆËãÄÜÁ¦£¬ÊôÓÚÖеµÌ⣮