Question

12£®ÒÑÖªA£¨x₀£¬0£©£¬B£¨0£¬y₀£©Á½µã·Ö±ðÔÚxÖáºÍyÖáÉÏÔ˶¯£¬ÇÒ|AB|=1£¬Èô¶¯µãP£¨x£¬y£©Âú×ã$\overrightarrow{OP}=2\overrightarrow{OA}+\sqrt{3}\overrightarrow{OB}$£®
£¨I£©Çó³ö¶¯µãPµÄ¹ì¼£¶ÔÓ¦ÇúÏßCµÄ±ê×¼·½³Ì£»
£¨¢ò£©Ò»Ìõ×ݽؾàΪ2µÄÖ±Ïßl₁ÓëÇúÏßC½»ÓÚP£¬QÁ½µã£¬ÈôÒÔPQÖ±¾¶µÄԲǡ¹ýÔ­µã£¬Çó³öÖ±Ïß·½³Ì£»
£¨¢ó£©Ö±Ïßl₂£ºx=ty+1ÓëÇúÏßC½»ÓÚA¡¢BÁ½µã£¬E£¨1£¬0£©£¬ÊÔÎÊ£ºµ±t±ä»¯Ê±£¬ÊÇ·ñ´æÔÚÒ»Ö±Ïßl₂£¬Ê¹¡÷ABEµÄÃæ»ýΪ$2\sqrt{3}$£¿Èô´æÔÚ£¬Çó³öÖ±Ïßl₂µÄ·½³Ì£»Èô²»´æÔÚ£¬ËµÃ÷ÀíÓÉ£®

Answer 1

·ÖÎö £¨¢ñ£©¸ù¾ÝÏòÁ¿µÄ×ø±êÔËË㣬ÒÔ¼°|AB|=1£¬µÃµ½ÍÖÔ²µÄ±ê×¼·½³ÌΪ$\frac{x^2}{4}+\frac{y^2}{3}=1$£®
£¨¢ò£©Ö±Ïßl₁бÂʱشæÔÚ£¬ÇÒ×ݽؾàΪ2£¬¸ù¾ÝÖ±ÏßÓëÍÖÔ²µÄλÖùØϵ£¬¼´¿ÉÇó³ökµÄÖµ£¬ÎÊÌâµÃÒÔ½â¾ö£®
£¨¢ó£©¸ù¾ÝÖ±ÏߺÍÍÖÔ²¶îλÖùØϵ£¬ÒÔ¼°Èý½ÇÐεÄÃæ»ý¹«Ê½µÃµ½S_¡÷ABE=$\frac{12\sqrt{{t}^{2}+1}}{3{t}^{2}+4}$£¬Áî=$\frac{12\sqrt{{t}^{2}+1}}{3{t}^{2}+4}$=2$\sqrt{3}$£¬Ôò${t^2}=-\frac{2}{3}$²»³ÉÁ¢£¬ÎÊÌâµÃÒÔ½â¾ö£®

½â´ð ½â£º£¨¢ñ£© ÒòΪ$\overrightarrow{OP}=2\overrightarrow{OA}+\sqrt{3}\overrightarrow{OB}$£¬
¼´$£¨x£¬y£©=2£¨{x_0}£¬0£©+\sqrt{3}£¨0£¬{y_0}£©=£¨2{x_0}£¬\sqrt{3}{y_0}£©$£¬
ËùÒÔ$x=2{x_0}£¬y=\sqrt{3}{y_0}$£¬
ËùÒÔ${x_0}=\frac{1}{2}x£¬{y_0}=\frac{{\sqrt{3}}}{3}y$
ÓÖÒòΪ|AB|=1£¬ËùÒÔ${x_0}^2+{y_0}^2=1$£¬
¼´£º${£¨\frac{1}{2}x£©^2}+{£¨\frac{{\sqrt{3}}}{3}y£©^2}=1$£¬
¼´$\frac{x^2}{4}+\frac{y^2}{3}=1$£¬
ËùÒÔÍÖÔ²µÄ±ê×¼·½³ÌΪ$\frac{x^2}{4}+\frac{y^2}{3}=1$£®
£¨¢ò£© Ö±Ïßl₁бÂʱشæÔÚ£¬ÇÒ×ݽؾàΪ2£¬ÉèÖ±ÏßΪy=kx+2ÁªÁ¢Ö±Ïßl₁ºÍÍÖÔ²·½³Ì$\left\{\begin{array}{l}y=kx+2\\ \frac{x^2}{4}+\frac{y^2}{3}=1\end{array}\right.$£¬
µÃ£º£¨3+4k²£©x²+16kx+4=0£¬
ÓÉ¡÷£¾0£¬µÃ${k^2}£¾\frac{1}{4}$£¨*£©£¬
ÉèP£¨x₁£¬y₁£©£¬Q£¨x₂£¬y₂£©£¬
Ôò${x_1}+{x_2}=-\frac{16k}{{3+4{k^2}}}£¬{x_1}{x_2}=\frac{4}{{3+4{k^2}}}$  £¨1£©
ÒÔPQÖ±¾¶µÄԲǡ¹ýÔ­µã£¬
ËùÒÔOP¡ÍOQ£¬$\overrightarrow{OP}•\overrightarrow{OQ}=0$£¬
¼´x₁x₂+y₁y₂=0£¬
Ò²¼´x₁x₂+£¨kx₁+2£©£¨kx₂+2£©=0£¬
¼´£¨1+k²£©x₁x₂+2k£¨x₁+x₂£©+4=0£¬
½«£¨1£©Ê½´úÈ룬µÃ$\frac{4£¨1+{k}^{2}£©}{3+4{k}^{2}}$-$\frac{32{k}^{2}}{3+4{k}^{2}}$+4=0£¬
¼´4£¨1+k²£©-32k²+4£¨3+4k²£©=0£¬
½âµÃ${k^2}=\frac{4}{3}$£¬Âú×㣨*£©Ê½£¬
ËùÒÔ$k=¡À\frac{{2\sqrt{3}}}{3}$£®
ËùÒÔÖ±Ïß·½³ÌΪy=¡À$\frac{2\sqrt{3}}{3}$x+2
£¨¢ó£©ÓÉ·½³Ì×é$\left\{\begin{array}{l}x=ty+1\\ \frac{x^2}{4}+\frac{y^2}{3}=1\end{array}\right.$£¬µÃ£¨3t²+4£©y²+6ty-9=0£¨*£©
ÉèA£¨x₁£¬y₁£©£¬B£¨x₂£¬y₂£©£¬
Ôò${y_1}+{y_2}=-\frac{6t}{{3{t^2}+4}}£¬{y_1}•{y_2}=-\frac{9}{{3{t^2}+4}}£¼0$
ËùÒÔ$\left|{{y_1}-{y_2}}\right|=\sqrt{{{£¨{{y_1}+{y_2}}£©}^2}-4{y_1}{y_2}}=\sqrt{{{£¨-\frac{6t}{{3{t^2}+4}}£©}^2}-4£¨-\frac{9}{{3{t^2}+4}}£©}=\frac{{12\sqrt{{t^2}+1}}}{{3{t^2}+4}}$£¬
ÒòΪֱÏßl£ºx=ty+1¹ýµãF£¨1£¬0£©£¬
ËùÒÔS_¡÷ABE=$\frac{1}{2}$|EF|•|y₁-y₂|=$\frac{1}{2}$¡Á2¡Á$\frac{12\sqrt{{t}^{2}+1}}{3{t}^{2}+4}$=$\frac{12\sqrt{{t}^{2}+1}}{3{t}^{2}+4}$
Áî=$\frac{12\sqrt{{t}^{2}+1}}{3{t}^{2}+4}$=2$\sqrt{3}$£¬Ôò${t^2}=-\frac{2}{3}$²»³ÉÁ¢
¹Ê²»´æÔÚÖ±ÏßlÂú×ãÌâÒ⣮

µãÆÀ ±¾Ì⿼²éÏòÁ¿µÄ×ø±êÔËËãÒÔ¼°ÍÖÔ²µÄ±ê×¼·½³ÌºÍÖ±ÏߺÍÍÖÔ²µÄλÖùØϵ£¬ÅàÑøÁËѧÉúµÄת»¯ÄÜÁ¦£¬ÔËËãÄÜÁ¦£¬ÊôÓÚÖеµÌ⣮