Question

13£®ÒÑÖªÍÖÔ²$C£º\frac{x^2}{a^2}+\frac{y^2}{b^2}=1£¨a£¾b£¾0£©$¹ýµã$A£¨-\frac{{\sqrt{2}}}{2}£¬\frac{{\sqrt{3}}}{2}£©$£¬ÇÒ¶ÌÖáÁ½¸ö¶¥µãÓëÒ»¸ö½¹µãÇ¡ºÃΪֱ½ÇÈý½ÇÐΣ®
£¨1£©ÇóÍÖÔ²CµÄ±ê×¼·½³Ì£»
£¨2£©ÊÇ·ñ´æÔÚÒÔÔ­µãΪԲÐĵÄÔ²£¬Ê¹µÃ¸ÃÔ²µÄÈÎÒâÒ»ÌõÇÐÏßÓëÍÖÔ²CºãÓÐÁ½¸ö½»µãP£¬Q£¬ÇÒ$\overrightarrow{OP}¡Í\overrightarrow{OQ}$£¿Èô´æÔÚ£¬Çó³ö¸ÃÔ²µÄ·½³Ì£»Èô²»´æÔÚ£¬Çë˵Ã÷ÀíÓÉ£®

Answer 1

·ÖÎö £¨1£©ÓÉÌâÒâµÃ£º$a=\sqrt{2}b$£¬$\frac{1}{2{b}^{2}}+\frac{3}{4{b}^{2}}$=1£¬ÓÉ´ËÄÜÇó³öÍÖÔ²CµÄ·½³Ì£®
£¨2£©¼ÙÉèÂú×ãÌõ¼þµÄÔ²´æÔÚ£¬Æä·½³ÌΪ£ºx²+y²=r²£¨0£¼r£¼1£©£¬ÉèÖ±Ïß·½³ÌΪy=kx+m£¬¶þÕßÁªÁ¢£¬µÃ£º£¨1+2k²£©x²+4kmx+2m²-2=0£¬ÓÉ´ËÀûÓÃΤ´ï¶¨Àí¡¢ÏòÁ¿´¹Ö±¡¢Ö±ÏßÓëÔ²ÏàÇУ¬½áºÏÒÑÖªÄÜÇó³ö´æÔÚÔ²ÐÄÔÚÔ­µãµÄÔ²Âú×ãÌâÒ⣮

½â´ð ½â£º£¨1£©¡ßÍÖÔ²$C£º\frac{x^2}{a^2}+\frac{y^2}{b^2}=1£¨a£¾b£¾0£©$¹ýµã$A£¨-\frac{{\sqrt{2}}}{2}£¬\frac{{\sqrt{3}}}{2}£©$£¬ÇÒ¶ÌÖáÁ½¸ö¶¥µãÓëÒ»¸ö½¹µãÇ¡ºÃΪֱ½ÇÈý½ÇÐΣ¬
¡àÓÉÌâÒâµÃ£º$a=\sqrt{2}b$£¬$\frac{1}{2{b}^{2}}+\frac{3}{4{b}^{2}}$=1£¬
½âµÃa=$\sqrt{2}$£¬b=1£¬
¡àÍÖÔ²CµÄ·½³ÌΪ$\frac{x^2}{2}+{y^2}=1$£®¡­£¨5·Ö£©
£¨2£©¼ÙÉèÂú×ãÌõ¼þµÄÔ²´æÔÚ£¬Æä·½³ÌΪ£ºx²+y²=r²£¨0£¼r£¼1£©
µ±Ö±ÏßP£¬QµÄбÂÊ´æÔÚʱ£¬ÉèÖ±Ïß·½³ÌΪy=kx+m£¬
ÓÉ$\left\{{\begin{array}{l}{y=kx+m}\\{{x^2}+2{y^2}=2}\end{array}}\right.$£¬µÃ£º£¨1+2k²£©x²+4kmx+2m²-2=0£¬
ÁîP£¨x₁£¬y₁£©£¬Q£¨x₂£¬y₂£©£¬ÔòÓУº${x_1}+{x_2}=-\frac{4km}{{1+2{k^2}}}$£¬${x_1}{x_2}=\frac{{2{m^2}-2}}{{1+2{k^2}}}$¡­£¨8·Ö£©
¡ß$\overrightarrow{OP}$¡Í$\overrightarrow{OQ}$£¬¡à${x_1}{x_2}+{y_1}{y_2}=0⇒£¨1+{k^2}£©{x_1}{x_2}+km£¨{x_1}+{x_2}£©+{m^2}=0$£®
¡à$\frac{{£¨1+{k^2}£©£¨2{m^2}-2£©}}{{1+2{k^2}}}-\frac{{4{k^2}{m^2}}}{{1+2{k^2}}}+{m^2}=0$£¬¡à3m²=2k²+2£®¡­£¨10·Ö£©
¡ßÖ±ÏßPQÓëÔ²ÏàÇУ¬¡à${r^2}=\frac{m^2}{{1+{k^2}}}=\frac{2}{3}$£¬¡à´æÔÚÔ²${x^2}+{y^2}=\frac{2}{3}$
µ±Ö±ÏßPQµÄбÂʲ»´æÔÚʱ£¬Ò²ÊʺÏ${x^2}+{y^2}=\frac{2}{3}$£®
×ÛÉÏËùÊö£¬´æÔÚÔ²ÐÄÔÚÔ­µãµÄÔ²${x^2}+{y^2}=\frac{2}{3}$Âú×ãÌâÒ⣮¡­£¨12·Ö£©

µãÆÀ ±¾Ì⿼²éÍÖÔ²µÄ·½³ÌµÄÇ󷨣¬¿¼²éÂú×ãÌõ¼þµÄÔ²µÄ·½³ÌÊÇ·ñ´æÔÚµÄÅжÏÓëÇ󷨣¬ÊÇÖеµÌ⣬½âÌâʱҪÈÏÕæÉóÌ⣬עÒâΤ´ï¶¨Àí¡¢ÏòÁ¿´¹Ö±¡¢Ö±ÏßÓëÔ²ÏàÇС¢ÍÖÔ²ÐÔÖʵĺÏÀíÔËÓã®