Question

3£®ÔÚƽÃæÖ±½Ç×ø±êϵxOyÖУ¬ÒÑÖª°ë¾¶Îª2µÄÔ²C£¬Ô²ÐÄÔÚxÖáÕý°ëÖáÉÏ£¬ÇÒÓëÖ±Ïßx-$\sqrt{3}$y+2=0ÏàÇУ®
£¨1£©ÇóÔ²CµÄ·½³Ì£»
£¨2£©ÔÚÔ²CÉÏ£¬ÊÇ·ñ´æÔÚµãP£¬Âú×ã|PQ|=$\frac{{\sqrt{2}}}{2}$|PO|£¬ÆäÖУ¬µãQµÄ×ø±êÊÇQ£¨-1£¬0£©£®Èô´æÔÚ£¬Ö¸³öÓм¸¸öÕâÑùµÄµã£»Èô²»´æÔÚ£¬Çë˵Ã÷ÀíÓÉ£»
£¨3£©ÈôÔÚÔ²CÉÏ´æÔÚµãM£¨m£¬n£©£¬Ê¹µÃÖ±Ïßl£ºmx+ny=1ÓëÔ²O£ºx²+y²=1Ïཻ²»Í¬Á½µãA£¬B£¬ÇómµÄÈ¡Öµ·¶Î§£®²¢Çó³öʹµÃ¡÷OABµÄÃæ»ý×î´óµÄµãMµÄ×ø±ê¼°¶ÔÓ¦µÄ¡÷OABµÄÃæ»ý£®

Answer 1

·ÖÎö £¨1£©¸ù¾ÝÖ±ÏߺÍÔ²ÏàÇУ¬¸ù¾Ýµãµ½Ö±ÏߵľàÀëµÈÓڰ뾶£¬½¨Á¢·½³Ì½øÐÐÇó½â¼´¿É£»
£¨2£©¸ù¾Ý|PQ|=$\frac{{\sqrt{2}}}{2}$|PO|£¬½¨Á¢·½³Ì¹Øϵ£¬½øÐÐÅжϼ´¿É£»
£¨3£©¸ù¾ÝÖ±ÏߺÍÔ²ÏཻµÄÐÔÖÊ£¬½áºÏÈý½ÇÐεÄÃæ»ý¹«Ê½½øÐÐÇó½â¼´¿É£®

½â´ð ½â£º£¨1£©ÉèÔ²ÐÄÊÇ£¨a£¬0£©£¬£¨a£¾0£©£¬Ëüµ½Ö±Ïßx-$\sqrt{3}$y+2=0µÄ¾àÀëÊÇd=$\frac{|a+2|}{\sqrt{1+3}}$=2£¬
½âµÃa=2»òa=-6£¨ÉáÈ¥£©£¬ËùÒÔ£¬ËùÇóÔ²CµÄ·½³ÌÊÇ£¨x-2£©²+y²=4£®£¨4·Ö£©
£¨2£©¼ÙÉè´æÔÚÕâÑùµÄµãP£¨x£¬y£©£¬ÔòÓÉ$PA=\frac{{\sqrt{2}}}{2}PO$£¬µÃx²+y²+4x+2=0£®£¨6·Ö£©
¼´£¬µãPÔÚÔ²D£º£¨x+2£©²+y²=2ÉÏ£¬µãPÒ²ÔÚÔ²C£º£¨x-2£©²+y²=4ÉÏ£®
ÒòΪ$|{CD}|=4£¾{r_c}+{r_d}=2+\sqrt{2}$£¬ËùÒÔÔ²CÓëÔ²DÍâÀ룬ԲCÓëÔ²DûÓй«¹²µã£®
ËùÒÔ£¬²»´æÔÚµãPÂú×ãÌõ¼þ£®£¨8·Ö£©
£¨3£©´æÔÚ£¬ÀíÓÉÈçÏ£ºÒòΪµãM£¨m£¬n£©£¬ÔÚÔ²CÉÏ£¬ËùÒÔ£¨m-2£©²+n²=4£¬
¼´n²=4-£¨m-2£©²=4m-m²ÇÒ0¡Üm¡Ü4£®
ÒòΪԭµãµ½Ö±Ïßl£ºmx+ny=1µÄ¾àÀëh=$\frac{1}{\sqrt{{m}^{2}+{n}^{2}}}$=$\frac{1}{\sqrt{4m}}$£¼1£¬½âµÃ$\frac{1}{4}$£¼m¡Ü4   £¨10·Ö£©
¶ø|AB|=2$\sqrt{1-{h}^{2}}$£¬
ËùÒÔS_¡÷OAB=$\frac{1}{2}$|AB|h=$\sqrt{{h}^{2}-{h}^{4}}$=$\sqrt{\frac{1}{4m}-£¨\frac{1}{4m}£©^{2}}$=$\sqrt{-£¨\frac{1}{4m}-\frac{1}{2}£©^{2}+\frac{1}{4}}$£¬
ÒòΪ$\frac{1}{16}¡Ü$$\frac{1}{4m}$£¼1£¬ËùÒÔµ±$\frac{1}{4m}$=$\frac{1}{2}$£¬¼´m=$\frac{1}{2}$ʱ£¬S_¡÷OABÈ¡µÃ×î´óÖµ$\frac{1}{2}$£¬
´ËʱµãMµÄ×ø±êÊÇ£¨$\frac{1}{2}$£¬$\frac{\sqrt{7}}{2}$£©»ò£¨$\frac{1}{2}$£¬-$\frac{\sqrt{7}}{2}$£©£¬¡÷OABµÄÃæ»ýµÄ×î´óÖµÊÇ$\frac{1}{2}$£®£¨12·Ö£©

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÖ±ÏߺÍÔ²µÄλÖùØϵµÄÓ¦Ó㬸ù¾ÝÌõ¼þ½¨Á¢·½³ÌÊǽâ¾ö±¾ÌâµÄ¹Ø¼ü£®¿¼²éѧÉúµÄ¼ÆËãÄÜÁ¦£®