Question

11£®ÒÑÖªÍÖÔ²C£º$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}$=1£¨a£¾b£¾0£©µÄÀëÐÄÂÊΪ$\frac{\sqrt{3}}{2}$£¬ÍÖÔ²CµÄ³¤°ëÖ᳤Ϊ2£®
£¨1£©ÇóÍÖÔ²CµÄ·½³Ì£»
£¨2£©ÒÑÖªÖ±Ïßl£ºy=kx-$\sqrt{3}$ÓëÍÖÔ²C½»ÓÚA£¬BÁ½µã£¬ÊÇ·ñ´æÔÚʵÊýkʹµÃÒÔÏ߶ÎABΪֱ¾¶µÄԲǡºÃ¾­¹ý×ø±êÔ­µãO£¿Èô´æÔÚ£¬Çó³ökµÄÖµ£»Èô²»´æÔÚ£¬Çë˵Ã÷ÀíÓÉ£®

Answer 1

·ÖÎö £¨1£©ÉèÍÖÔ²µÄ½¹°ë¾àΪc£¬ÔòÓÉÌâÉ裬µÃ$\left\{{\begin{array}{l}{a=2}\\{\frac{c}{a}=\frac{{\sqrt{3}}}{2}}\end{array}}\right.$£¬Çó³öÍÖÔ²CµÄ¼¸ºÎÁ¿£¬È»ºóÇó½âÍÖÔ²·½³Ì£®
£¨2£©´æÔÚʵÊýkʹµÃÒÔÏ߶ÎABΪֱ¾¶µÄԲǡºÃ¾­¹ý×ø±êÔ­µãO£®ÀíÓÉÈçÏ£ºÉèµãA£¨x₁£¬y₁£©£¬B£¨x₂£¬y₂£©£¬½«Ö±ÏßlµÄ·½³Ì$y=kx-\sqrt{3}$´úÈë$\frac{x^2}{4}+{y^2}=1$£¬ÀûÓÃΤ´ï¶¨ÀíÒÔ¼°ÏòÁ¿µÄÊýÁ¿»ý£¬×ª»¯Çó½â¼´¿É£®

½â´ð £¨1£©ÉèÍÖÔ²µÄ½¹°ë¾àΪc£¬ÔòÓÉÌâÉ裬µÃ$\left\{{\begin{array}{l}{a=2}\\{\frac{c}{a}=\frac{{\sqrt{3}}}{2}}\end{array}}\right.$£¬½âµÃ$\left\{{\begin{array}{l}{a=2}\\{c=\sqrt{3}}\end{array}}\right.$£¬¡­£¨2·Ö£©
ËùÒÔb²=a²-c²=4-3=1£¬¹ÊËùÇóÍÖÔ²CµÄ·½³ÌΪ$\frac{x^2}{4}+{y^2}=1$£®¡­..£¨4·Ö£©
£¨2£©´æÔÚʵÊýkʹµÃÒÔÏ߶ÎABΪֱ¾¶µÄԲǡºÃ¾­¹ý×ø±êÔ­µãO£®ÀíÓÉÈçÏ£º
ÉèµãA£¨x₁£¬y₁£©£¬B£¨x₂£¬y₂£©£¬½«Ö±ÏßlµÄ·½³Ì$y=kx-\sqrt{3}$´úÈë$\frac{x^2}{4}+{y^2}=1$£¬
²¢ÕûÀí£¬µÃ$£¨1+4{k^2}£©{x^2}-8\sqrt{3}x+8=0$£®£¨*£©¡­£®£¨6·Ö£©
Ôò${x_1}+{x_2}=\frac{{8\sqrt{3}k}}{{1+4{k^2}}}$£¬${x_1}{x_2}=\frac{8}{{1+4{k^2}}}$£®¡­£¨8·Ö£©
ÒòΪÒÔÏ߶ÎABΪֱ¾¶µÄԲǡºÃ¾­¹ý×ø±êÔ­µãO£¬ËùÒÔ$\overrightarrow{OA}•\overrightarrow{OB}=0$£¬¼´x₁x₂+y₁y₂=0£®
ÓÖ${y_1}{y_2}={k^2}{x_1}{x_2}-\sqrt{3}k£¨{x_1}+{x_2}£©+3$£¬ÓÚÊÇ$\frac{8}{{1+4{k^2}}}-\frac{{4{k^2}-3}}{{1+4{k^2}}}=0$£¬¡­£®£¨10·Ö£©
½âµÃ$k=¡À\frac{{\sqrt{11}}}{2}$£¬¡­..£¨11·Ö£©
¾­¼ìÑéÖª£º´Ëʱ£¨*£©Ê½µÄ¡÷£¾0£¬·ûºÏÌâÒ⣮
ËùÒÔµ±$k=¡À\frac{{\sqrt{11}}}{2}$ʱ£¬ÒÔÏ߶ÎABΪֱ¾¶µÄԲǡºÃ¾­¹ý×ø±êÔ­µãO£®¡­£¨12·Ö£©

µãÆÀ ±¾Ì⿼²éÍÖÔ²·½³ÌµÄÇ󷨣¬Ö±ÏßÓëÍÖÔ²µÄλÖùØϵµÄ×ÛºÏÓ¦Ó㬿¼²é´æÔÚÐÔÎÊÌâµÄ´¦Àí·½·¨£¬¿¼²é¼ÆËãÄÜÁ¦£®