Question

12£®ÒÑÖªÍÖÔ²CµÄÖÐÐÄÔÚ×ø±êÔ­µãO£¬×ó¡¢ÓÒ½¹µã·Ö±ðΪF₁¡¢F₂£¬µãAʱÍÖÔ²CÉÏÈÎÒ»µã£¬ÇÒ|AF₁|•|AF₂|µÄ×î´óֵΪ3£¬ÒÔÍÖÔ²CµÄÓÒ½¹µãΪԲÐÄ£¬½¹¾àΪֱ¾¶µÄÔ²ÓëÖ±Ïßl₁£ºx+$\sqrt{3}$y+1=0ÏàÇУ®
£¨1£©ÇóÍÖÔ²CµÄ±ê×¼·½³Ì£»
£¨2£©²»¹ýÔ­µãµÄÖ±Ïßl₂ÓëÍÖÔ²C½»ÓÚP£¨x₁£¬y₁£©£¬Q£¨x₂£¬y₂£©Á½¸ö²»Í¬µã£¬ÒÔOP£¬OQΪÁÚ±ß×÷?OQNP£¬µ±?OQNPµÄÃæ»ýΪ$\sqrt{6}$ʱ£¬Ö¤Ã÷£º|ON|²+|PQ|²Îª¶¨Öµ£®

Answer 1

·ÖÎö £¨1£©ÉèÍÖÔ²µÄ±ê×¼·½³ÌΪ£º$\frac{{x}^{2}}{{a}^{2}}$+$\frac{{y}^{2}}{{b}^{2}}$=1£¨a£¾b£¾0£©£®Éè|AF₁|=m£¬|AF₂|=n£¬Ôòmn¡Ü3£¬ÓÉÓÚm+n=2a£¬ÀûÓûù±¾²»µÈʽµÄÐÔÖÊ£ºa²=3£®ÓÉÓÚÒÔÍÖÔ²CµÄÓÒ½¹µãΪԲÐÄ£¬½¹¾àΪֱ¾¶µÄÔ²ÓëÖ±Ïßl₁£ºx+$\sqrt{3}$y+1=0ÏàÇУ®¿ÉµÃ$\frac{|c+1|}{2}$=c£¬c£¾0£¬½âµÃc£¬¿ÉµÃb²=a²-c²£®
£¨2£©µ±Ö±Ïßl₂µÄбÂÊ´æÔÚʱ£¬ÉèÖ±Ïßl₂µÄ·½³ÌΪy=kx+m£¬Ï߶ÎPQµÄÖеãM£¨x₀£¬y₀£©£®ÓëÍÖÔ²·½³ÌÁªÁ¢»¯Îª£º£¨2+3k²£©x²+6kmx+3m²-6=0£¬ÀûÓøùÓëϵÊýµÄ¹Øϵ¼°ÆäÖеã×ø±ê¹«Ê½¿ÉµÃ£º|ON|²=4|OM|²=$4£¨{x}_{0}^{2}+{y}_{0}^{2}£©$£¬|PQ|²=£¨1+k²£©$[£¨{x}_{1}+{x}_{2}£©^{2}-4{x}_{1}{x}_{2}]$£®Ô­µãOµ½Ö±ÏßPQµÄ¾àÀëd=$\frac{|m|}{\sqrt{1+{k}^{2}}}$£¬Æ½ÐÐËıßÐÎOQNPµÄÃæ»ýS=2¡Á$\frac{1}{2}$|PQ|•d=|PQ|•$\frac{|m|}{\sqrt{1+{k}^{2}}}$=$\sqrt{6}$£¬¿ÉµÃ|PQ|²=$\frac{6£¨1+{k}^{2}£©}{{m}^{2}}$£®µÃµ½mÓëkµÄ¹Øϵ£¬´úÈë|ON|²+|PQ|²£¬¼´¿ÉÖ¤Ã÷£®µ±Ö±Ïßl₂µÄбÂʲ»´æÔÚʱ£¬Í¬Ñù³ÉÁ¢£®

½â´ð £¨1£©½â£ºÉèÍÖÔ²µÄ±ê×¼·½³ÌΪ£º$\frac{{x}^{2}}{{a}^{2}}$+$\frac{{y}^{2}}{{b}^{2}}$=1£¨a£¾b£¾0£©£®
Éè|AF₁|=m£¬|AF₂|=n£¬Ôòmn¡Ü3£¬
¡ßm+n=2a¡Ý2$\sqrt{mn}$£¬µ±ÇÒ½öµ±m=nʱȡµÈºÅ£¬a²=3£®
¡ßÒÔÍÖÔ²CµÄÓÒ½¹µãΪԲÐÄ£¬½¹¾àΪֱ¾¶µÄÔ²ÓëÖ±Ïßl₁£ºx+$\sqrt{3}$y+1=0ÏàÇУ®
¡à$\frac{|c+1|}{2}$=c£¬c£¾0£¬½âµÃc=1£¬
¡àb²=a²-c²=2£®
¡àÍÖÔ²CµÄ±ê×¼·½³ÌΪ$\frac{{x}^{2}}{3}+\frac{{y}^{2}}{2}$=1£®
£¨2£©Ö¤Ã÷£ºµ±Ö±Ïßl₂µÄбÂÊ´æÔÚʱ£¬ÉèÖ±Ïßl₂µÄ·½³ÌΪy=kx+m£¬Ï߶ÎPQµÄÖеãM£¨x₀£¬y₀£©£®
ÁªÁ¢$\left\{\begin{array}{l}{y=kx+m}\\{\frac{{x}^{2}}{3}+\frac{{y}^{2}}{2}=1}\end{array}\right.$£¬»¯Îª£º£¨2+3k²£©x²+6kmx+3m²-6=0£¬
¡÷£¾0£¬¿ÉµÃx₁+x₂=-$\frac{6km}{2+3{k}^{2}}$£¬x₁x₂=$\frac{3{m}^{2}-6}{2+3{k}^{2}}$£®
¡àx₀=$\frac{{x}_{1}+{x}_{2}}{2}$=$\frac{-3km}{2+3{k}^{2}}$£¬y₀=kx₀+m=$\frac{2m}{2+3{k}^{2}}$£®
¡à|ON|²=4|OM|²=$4£¨{x}_{0}^{2}+{y}_{0}^{2}£©$=$\frac{4{m}^{2}£¨9{k}^{2}+4£©}{£¨2+3{k}^{2}£©^{2}}$£®
|PQ|²=£¨1+k²£©$[£¨{x}_{1}+{x}_{2}£©^{2}-4{x}_{1}{x}_{2}]$=£¨1+k²£©$[\frac{36{k}^{2}{m}^{2}}{£¨2+3{k}^{2}£©^{2}}-\frac{4£¨3{m}^{2}-6£©}{2+3{k}^{2}}]$=$\frac{12£¨4+6{k}^{2}-2{m}^{2}£©}{£¨2+3{k}^{2}£©^{2}}$¡Á£¨1+k²£©£®
¡à|ON|²+|PQ|²=$\frac{12{m}^{2}{k}^{2}+120{k}^{2}+72{k}^{4}-8{m}^{2}+48}{£¨2+3{k}^{2}£©^{2}}$£®
Ô­µãOµ½Ö±ÏßPQµÄ¾àÀëd=$\frac{|m|}{\sqrt{1+{k}^{2}}}$£¬
ƽÐÐËıßÐÎOQNPµÄÃæ»ýS=2¡Á$\frac{1}{2}$|PQ|•d=|PQ|•$\frac{|m|}{\sqrt{1+{k}^{2}}}$=$\sqrt{6}$£¬
¡à|PQ|²=$\frac{6£¨1+{k}^{2}£©}{{m}^{2}}$£®
¡à$\frac{12£¨4+6{k}^{2}-2{m}^{2}£©}{£¨2+3{k}^{2}£©^{2}}$¡Á£¨1+k²£©=$\frac{6£¨1+{k}^{2}£©}{{m}^{2}}$£¬»¯Îª£º2m²=2+3k²£®
´úÈë|ON|²+|PQ|²=$\frac{£¨2+3{k}^{2}£©£¨6{k}^{2}-4£©+120{k}^{2}+72{k}^{4}+48}{£¨2+3{k}^{2}£©^{2}}$=10£®
¡à|ON|²+|PQ|²Îª¶¨Öµ10£®
µ±Ö±Ïßl₂µÄбÂʲ»´æÔÚʱ£¬Í¬Ñù³ÉÁ¢£®
×ÛÉϿɵãº|ON|²+|PQ|²Îª¶¨Öµ10£®

µãÆÀ ±¾Ì⿼²éÁËÍÖÔ²µÄ¶¨Òå±ê×¼·½³Ì¼°ÆäÐÔÖÊ¡¢Ö±ÏßÓëÍÖÔ²ÏཻÎÊÌâ¡¢Ö±ÏßÓëÔ²ÏàÇÐÎÊÌâ¡¢Ò»Ôª¶þ´Î·½³ÌµÄ¸ùÓëϵÊýµÄ¹Øϵ¡¢Æ½ÐÐËıßÐεÄÃæ»ý¼ÆË㹫ʽ¡¢µãµ½Ö±ÏߵľàÀ빫ʽ¹«Ê½¡¢ÏÒ³¤¹«Ê½¡¢»ù±¾²»µÈʽµÄÐÔÖÊ£¬¿¼²éÁËÍÆÀíÄÜÁ¦Óë¼ÆËãÄÜÁ¦£¬ÊôÓÚÄÑÌ⣮