Question

4£®Èçͼ£¬¡÷ABCÊÇÒÔBCΪµ×±ßµÄµÈÑüÈý½ÇÐΣ¬µãA¡¢C·Ö±ðÊÇÒ»´Îº¯Êý$y=-\frac{3}{4}x+3$µÄͼÏóÓëyÖá¡¢xÖáµÄ½»µã£¬µãBÔÚ¶þ´Îº¯Êý$y=\frac{1}{8}{x^2}+bx+c$µÄͼÏóÉÏ£¬ÇҸöþ´Îº¯ÊýͼÏóÉÏ´æÔÚÒ»µãDʹËıßÐÎABCDÄܹ¹³ÉƽÐÐËıßÐΣ®
£¨1£©ÊÔÇób¡¢cµÄÖµ£¬²¢Ð´³ö¸Ã¶þ´Îº¯ÊýµÄ½âÎöʽ£»
£¨2£©¶¯µãP´ÓAµ½D£¬Í¬Ê±¶¯µãQ´ÓCµ½A¶¼ÒÔÿÃë1¸öµ¥Î»µÄËÙ¶ÈÔ˶¯£¬ÎÊ£º
¢Ùµ±PÔ˶¯µ½ºÎ´¦Ê±£¬¡÷APQÊÇÖ±½ÇÈý½ÇÐΣ¿
¢Úµ±PÔ˶¯µ½ºÎ´¦Ê±£¬ËıßÐÎPDCQµÄÃæ»ý×îС£¿´ËʱËıßÐÎPDCQµÄÃæ»ýÊǶàÉÙ£¿

Answer 1

·ÖÎö £¨1£©Ê×ÏȵóöB£¬Dµã×ø±ê£¬ÔÙÀûÓôý¶¨ÏµÊý·¨Çó³ö¶þ´Îº¯Êý½âÎöʽ¼´¿É£»
£¨2£©¢Ù·Ö±ðÀûÓõ±PQ¡ÍACʱ£¬µ±QP¡ÍADʱ£¬½áºÏ¹´¹É¶¨ÀíÇó³ötµÄÖµ¼´¿É£»
¢Ú¹ýµãQ×÷QH¡ÍAD£¬´¹×ãΪH£®ÓÉÓÚS_¡÷APQ=$\frac{1}{2}$AP•QH=$\frac{1}{2}$AP•AQsin¡ÏPAQ£¬S_¡÷ACD=$\frac{1}{2}$AD•OA£¬ËùÒÔS_{ËıßÐÎPDCQ}=S_¡÷ACD-S_¡÷APQÇó³ö×îÖµ¼´¿É£®

½â´ð ½â£º£¨1£©ÓÉ$y=-\frac{3}{4}x+3$£¬µÃA£¨0£¬3£©£¬C£¨4£¬0£©£®
ÓÉÓÚB¡¢C¹ØÓÚOA¶Ô³Æ£¬ËùÒÔB£¨-4£¬0£©£¬BC=8£®
ÒòΪAD¡ÎBC£¬AD=BC£¬ËùÒÔD£¨8£¬3£©£®
½«B£¨-4£¬0£©¡¢D£¨8£¬3£©·Ö±ð´úÈë$y=\frac{1}{8}{x^2}+bx+c$£¬
µÃ$\left\{\begin{array}{l}2-4b+c=0\\ 8+8b+c=3.\end{array}\right.$
½âµÃ $b=-\frac{1}{4}$£¬c=-3£®
ËùÒԸöþ´Îº¯ÊýµÄ½âÎöʽΪ$y=\frac{1}{8}{x^2}-\frac{1}{4}x-3$£®

£¨2£©¢ÙÉèµãP¡¢QÔ˶¯µÄʱ¼äΪt£®
ÔÚ¡÷APQÖУ¬AP=t£¬AQ=AC-CQ=5-t£¬cos¡ÏPAQ=cos¡ÏACO=$\frac{4}{5}$£®
µ±PQ¡ÍACʱ£¬$\frac{AQ}{AP}=\frac{4}{5}$£®
ËùÒÔ$\frac{5-t}{t}$=$\frac{4}{5}$£®
½âµÃt=$\frac{25}{9}$£¬¼´AP=$\frac{25}{9}$£®£¨Èçͼ1Ëùʾ£©
µ±QP¡ÍADʱ£®Õâʱ$\frac{AP}{AQ}=\frac{4}{5}$£¬
ËùÒÔ$\frac{t}{5-t}=\frac{4}{5}$£®
½âµÃ$t=\frac{20}{9}$£¨Èçͼ2Ëùʾ£©£®

¢ÚÈçͼ3£¬¹ýµãQ×÷QH¡ÍAD£¬´¹×ãΪH£®
ÓÉÓÚS_¡÷APQ=$\frac{1}{2}AP•QH=\frac{1}{2}AP•AQsin¡ÏPAQ=\frac{1}{2}t£¨5-t£©¡Á\frac{3}{5}=-\frac{3}{10}{t^2}+\frac{3}{2}t$

S_¡÷ACD=$\frac{1}{2}AD•OA=\frac{1}{2}¡Á8¡Á3=12$£¬
ËùÒÔS_{ËıßÐÎPDCQ}=S_¡÷ACD-S_¡÷APQ=$12-£¨-\frac{3}{10}{t^2}+\frac{3}{2}t£©=\frac{3}{10}{£¨t-\frac{5}{2}£©^2}+\frac{81}{8}$£®
ËùÒÔµ±AP=$\frac{5}{2}$ʱ£¬ËıßÐÎPDCQÃæ»ýµÄ×îСֵÊÇ$\frac{81}{8}$£®

µãÆÀ ´ËÌâÖ÷Òª¿¼²éÁ˶þ´Îº¯Êý×ÛºÏÒÔ¼°Ö±½ÇÈý½ÇÐεÄÐÔÖÊÒÔ¼°¶þ´Îº¯Êý×îÖµÇ󷨵È֪ʶ£¬ÀûÓÃÊýÐνáºÏÒÔ¼°·ÖÀàÌÖÂ۵óöÊǽâÌâ¹Ø¼ü£®