Question

8£®ÒÑÖªÍÖÔ²µÄÖÐÐÄÊÇ×ø±êÔ­µãO£¬½¹µãÔÚxÖáÉÏ£¬ÀëÐÄÂÊΪ$\frac{\sqrt{2}}{2}$£¬¶ÌÖ᳤Ϊ2£¬¶¨µãA£¨2£¬0£©£¬µãPÔÚÒÑÖªÍÖÔ²ÉÏ£¬¶¯µãQÂú×ã$\overrightarrow{OQ}$=$\overrightarrow{OA}$+$\overrightarrow{OP}$£®
£¨1£©Ç󶯵ãQµÄ¹ì¼£·½³Ì£»
£¨2£©¹ýÍÖÔ²ÓÒ½¹µãFµÄÖ±ÏßÓëÍÖÔ²½»ÓÚµãM£¬N£¬µ±|MN|×îСʱ£¬Çó¡÷AMNµÄÃæ»ý£®

Answer 1

·ÖÎö £¨1£©ÒÀÌâÒâÔËÓÃÀëÐÄÂʹ«Ê½ºÍa£¬b£¬cµÄ¹Øϵ£¬Ò×ÇóµÃÍÖÔ²µÄ±ê×¼·½³Ì£®Éè³öµãQµÄ×ø±ê£¬½áºÏ¶¨µãA£¨2£¬0£©£¬PÔÚÒÑÖªÍÖÔ²ÉÏÒÔ¼°¹Øϵʽ$\overrightarrow{OQ}=\overrightarrow{OA}+\overrightarrow{OP}$£®¿ÉÒÔÇ󶯵ãQ¹ì¼£·½³Ì£»
£¨2£©ÉèµãAµ½Ö±ÏßMNµÄ¾àÀëΪd£¬Ôò¡÷AMNµÄÃæ»ý$S=\frac{1}{2}\left|MN\right|d$£¬ÆäÖÐ|MN|¿ÉÒÔÀûÓÃÏÒ³¤¹«Ê½ÇóµÃ£¬ÀûÓú¯Êý»ò»ù±¾²»µÈʽÇó×îÖµ£®

½â´ð ½â£º£¨1£©ÉèÍÖÔ²µÄ±ê×¼·½³ÌΪ$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1£¨a£¾b£¾0£©$£¬
ÓÉÌâÒâ¿ÉÖª$\left\{\begin{array}{l}e=\frac{\sqrt{2}}{2}\\ b=1\\{c}^{2}={a}^{2}-{b}^{2}\end{array}\right.$£¬¼´$\left\{\begin{array}{l}\frac{c}{a}=\frac{\sqrt{2}}{2}\\ b=1\\{c}^{2}={a}^{2}-1\end{array}\right.$½âµÃ$\left\{\begin{array}{l}a=\sqrt{2}\\ b=1\\ c=1\end{array}\right.$£¬
¹ÊÍÖÔ²µÄ±ê×¼·½³ÌΪ$\frac{{x}^{2}}{2}+{y}^{2}=1$£»
ÉèQ£¨x£¬y£©£¬P£¨x₁£¬y₁£©£¬ÒòΪA£¨2£¬0£©£¬
ËùÒÔ$\overrightarrow{OP}=\overrightarrow{OA}-\overrightarrow{OQ}=£¨2-x£¬-y£©=£¨{x}_{1}£¬{y}_{1}£©$£¬
ËùÒÔx₁=2-x£¬y₁=-y£¬
ÓÖµãPÔÚÒÑÖªÍÖÔ²ÉÏ£¬
¹Ê$\frac{{£¨x-2£©}^{2}}{2}+{y}^{2}=1$Ϊ¶¯µãQµÄ¹ì¼£·½³Ì£®
£¨2£©ÍÖÔ²µÄÓÒ½¹µãF£¨1£¬0£©£¬ÉèÖ±ÏßMNµÄ·½³ÌÊÇx=my+1£¬Óë$\frac{{x}^{2}}{2}+{y}^{2}=1$ÁªÁ¢£¬
¿ÉµÃ£¨m²+2£©y²+2my-1=0£¬
ÉèM£¨x₁£¬y₁£©£¬N£¨x₂£¬y₂£©£¬Ôòx₁=my₁+1£¬x₂=my₁₂+1£¬
ÓÉÌâÒây₁£¬y₂Âú×ã·½³Ì£¨m²+2£©y²+2my-1=0£¬
¡÷=4m²+4£¨m²+2£©£¾0¼´m²+1£¾0£¬
Ôò·½³Ì¸ùÓëϵÊýµÄ¹Øϵ¿ÉµÃ£º${y}_{1}+{y}_{2}=-\frac{2m}{£¨{m}^{2}+2£©}{£¬y}_{1}{y}_{2}=\frac{-1}{£¨{m}^{2}+2£©}$
¼´ÓÐ$\left|MN\right|=\sqrt{{£¨{x}_{1}-{x}_{2}£©}^{2}+{£¨{y}_{1}-{y}_{2}£©}^{2}}=\sqrt{£¨{m}^{2}+1£©}|{y}_{1}-{y}_{2}|$£¬
ÓÖ $\left|{y}_{1}-{y}_{2}\right|=\sqrt{{£¨{y}_{1}-{y}_{2}£©}^{2}}=\sqrt{{£¨{y}_{1}+{y}_{2}£©}^{2}-4{y}_{1}{y}_{2}}$=$\sqrt{{£¨\frac{-2m}{{m}^{2}+2}£©}^{2}-4¡Á£¨\frac{-1}{{m}^{2}+2}£©}$=$\frac{2\sqrt{2£¨{m}^{2}+1£©}}{{m}^{2}+2}$
Ôò$\left|MN\right|=\frac{2\sqrt{2}£¨{m}^{2}+1£©}{{m}^{2}+2}$£¬
µãA£¨2£¬0£©µ½Ö±ÏßMNµÄ¾àÀë$d=\frac{1}{\sqrt{{m}^{2}+1}}$£¬
ÓÚÊÇ¡÷AMNµÄÃæ»ý$S=\frac{1}{2}\left|MN\right|d=\frac{\sqrt{2£¨{m}^{2}+1£©}}{{m}^{2}+2}$
=$\sqrt{\frac{2}{£¨{m}^{2}+1£©+\frac{1}{{m}^{2}+1}+2}}¡Ü\sqrt{\frac{2}{2\sqrt{£¨{m}^{2}+1£©¡Á£¨\frac{1}{{m}^{2}+1}}£©+2}}=\frac{\sqrt{2}}{2}$£¬
µ±ÇÒ½öµ±${m}^{2}+1=\frac{1}{{m}^{2}+1}$£¬¼´m=0ʱȡµ½µÈºÅ£®
¹Ê¡÷AMNµÄÃæ»ýµÄ×î´óÖµÊÇ$\frac{\sqrt{2}}{2}$£®

µãÆÀ ±¾Ì⿼²éÍÖÔ²µÄ·½³ÌµÄÇ󷨺ÍÔËÓ㬿¼²é¹ì¼£·½³ÌµÄÇ󷨣¬×¢ÒâÔËÓôúÈë·¨£¬Í¬Ê±¿¼²éÖ±ÏߺÍÍÖÔ²·½³ÌÁªÁ¢£¬ÔËÓÃΤ´ï¶¨ÀíºÍÅбðʽ£¬ÒÔ¼°ÏÒ³¤¹«Ê½£¬¿¼²éÔËË㻯¼òÄÜÁ¦£¬ÊôÓÚÖеµÌ⣮