Question

3£®ÉèA£¬B·Ö±ðÊÇÖ±Ïßy=$\frac{{2\sqrt{5}}}{5}$xºÍy=-$\frac{{2\sqrt{5}}}{5}$xÉϵĶ¯µã£¬ÇÒ|AB|=2$\sqrt{5}$£¬ÉèOΪ×ø±êÔ­µã£¬¶¯µãPÂú×ã$\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{OB}$£®
£¨¢ñ£©Ç󶯵ãPµÄ¹ì¼£·½³Ì£»
£¨¢ò£©Ð±ÂÊΪ1²»¾­¹ýÔ­µãO£¬ÇÒÓ붯µãPµÄ¹ì¼£ÏཻÓÚC£¬DÁ½µã£¬MΪÏ߶ÎCDµÄÖе㣬ֱÏßCDÓëÖ±ÏßOMÄÜ·ñ´¹Ö±£¿Ö¤Ã÷ÄãµÄ½áÂÛ£®

Answer 1

·ÖÎö £¨¢ñ£©Éè³öA£¬B×ø±ê£¬PµÄ×ø±ê£¬ÀûÓÃÏòÁ¿¹Øϵ£¬|AB|¾àÀë¼´¿ÉÇ󶯵ãPµÄ¹ì¼£·½³Ì£»
£¨¢ò£©Ö±ÏßCDÓëÖ±ÏßOM²»´¹Ö±£®ÉèC£¨x₃£¬y₃£©£¬D£¨x₄£¬y₄£©£¬ÀûÓÃƽ·½²î·¨×ª»¯Çó½â¼´¿É£®

½â´ð ½â£º£¨¢ñ£©Éè$A£¨{x_1}£¬\frac{{2\sqrt{5}}}{5}{x_1}£©£¬B£¨{x_2}£¬-\frac{{2\sqrt{5}}}{5}{x_2}£©£¬P£¨x£¬y£©$£¬¡­£¨1·Ö£©
¡ß$\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{OB}$£¬¡à$x={x_1}+{x_2}£¬y=\frac{{2\sqrt{5}}}{5}£¨{x_1}-{x_2}£©$£®¡­£¨3·Ö£©
¡ß$|AB|=2\sqrt{5}$£¬¡à$20={£¨{x_1}-{x_2}£©^2}+{£¨\frac{{2\sqrt{5}}}{5}{x_1}+\frac{{2\sqrt{5}}}{5}{x_2}£©^2}$£¬¡­£¨5·Ö£©£¬
$20=\frac{5}{4}{y^2}+\frac{4}{5}{x^2}$£¬
¡à¶¯µãPµÄ¹ì¼£·½³Ì$\frac{x^2}{25}+\frac{y^2}{16}=1$£®¡­£¨6·Ö£©
£¨¢ò£©Ö±ÏßCDÓëÖ±ÏßOM²»´¹Ö±£®
ÉèC£¨x₃£¬y₃£©£¬D£¨x₄£¬y₄£©£¬$\left\{\begin{array}{l}\frac{x_3^2}{25}+\frac{y_3^2}{16}=1\\ \frac{x_4^2}{25}+\frac{y_4^2}{16}=1\end{array}\right.$¡­£¨8·Ö£©
$\frac{{£¨{x_3}-{x_4}£©£¨{x_3}+{x_4}£©}}{25}+\frac{{£¨{y_3}-{y_4}£©£¨{y_3}+{y_4}£©}}{16}=0$£¬¡­£¨10·Ö£©
¡ßÖ±ÏßCDµÄбÂÊΪ1£¬
¡à$\frac{{£¨{y_3}+{y_4}£©}}{{£¨{x_3}+{x_4}£©}}=-\frac{16}{25}$£¬¡­£¨11·Ö£©
¡àÖ±ÏßOMµÄбÂÊΪ$-\frac{16}{25}$£¬
¡àÖ±ÏßCDÓëÖ±ÏßOM²»´¹Ö±£®¡­£¨12·Ö£©

µãÆÀ ±¾Ì⿼²é¹ì¼£·½³ÌµÄÇ󷨣¬Ö±ÏßÓëԲ׶ÇúÏßλÖùØϵµÄ×ÛºÏÓ¦Ó㬿¼²éת»¯Ë¼ÏëÒÔ¼°¼ÆËãÄÜÁ¦£®