Question

9£®ÒÑÖªÍÖÔ²C£º$\frac{{x}^{2}}{{a}^{2}}$+$\frac{{y}^{2}}{{b}^{2}}$=1µÄÀëÐÄÂÊe=$\frac{1}{2}$£¬F£¨1£¬0£©£¬ÊÇÍÖÔ²CµÄÓÒ½¹µã£¬Èô²»¾­¹ýÔ­µãOµÄÖ±Ïßl£ºy=kx+m£¨k£¾0£©ÓëÍÖÔ²CÏཻÓÚ²»Í¬µÄÁ½µãA¡¢B£¬¼ÇÖ±ÏßOA£¬OBµÄбÂÊ·Ö±ðΪk₁£¬k₂£¬ÇÒk₁•k₂=k²£®
£¨¢ñ£©ÇóÍÖÔ²CµÄ·½³Ì£»
£¨¢ò£©ÇóÖ¤£ºÖ±ÏßABµÄбÂÊΪ¶¨Öµ£¬²¢Çó¡÷AOBÃæ»ýµÄ×î´óÖµ£®

Answer 1

·ÖÎö £¨¢ñ£©ÓÉÌõ¼þÀûÓÃÍÖÔ²µÄÐÔÖÊÇó³öa¡¢bµÄÖµ£¬¿ÉµÃÍÖÔ²CµÄ·½³Ì£®
£¨¢ò£©°ÑÖ±Ïßl£ºy=kx+m£¨k£¾0£©´úÈëÍÖÔ²µÄ·½³Ì£¬ÀûÓÃΤ´ï¶¨Àí¡¢µãµ½Ö±ÏߵľàÀ빫ʽ¡¢ÏÒ³¤¹«Ê½Çó³öS_¡÷AOB=$\frac{1}{2}$•AB•d=$\frac{1}{2}$•$\sqrt{\frac{4£¨6{-m}^{2}£©{•m}^{2}}{3}}$¡Ü$\sqrt{3}$£¬´Ó¶øÖ¤µÃ½áÂÛ£®

½â´ð ½â£º£¨¢ñ£©ÓÉÌâÒâ¿ÉµÃe=$\frac{c}{a}$=$\frac{1}{2}$£¬c=1£¬¡àa=2£¬b=$\sqrt{3}$£¬
¡àÍÖÔ²CµÄ·½³ÌΪ $\frac{{x}^{2}}{4}$+$\frac{{y}^{2}}{3}$=1£®
£¨¢ò£©Ö¤Ã÷£ºÉèµãA£¨x₁£¬y₁£©¡¢B£¨x₂£¬y₂£©£¬°ÑÖ±Ïßl£ºy=kx+m£¨k£¾0£©´úÈëÍÖÔ² $\frac{{x}^{2}}{4}$+$\frac{{y}^{2}}{3}$=1£¬
¿ÉµÃ £¨4k²+3£©x²+8kmx+4m²-12=0£®
ÓÉ¡÷=48£¨3-m²+4k²£©£¾0£¬x₁+x₂=-$\frac{8km}{{4k}^{2}+3}$£¬x₁•x₂=$\frac{{4m}^{2}-12}{{4k}^{2}+3}$£®
ÓÉk₁•k₂=$\frac{£¨{kx}_{1}+m£©•£¨{kx}_{2}+m£©}{{x}_{1}{•x}_{2}}$=k²+$\frac{km{£¨x}_{1}{+x}_{2}£©{+m}^{2}}{{x}_{1}{•x}_{2}}$=k²£®
¿ÉµÃkm£¨x₁•x₂ £©+m²=0£¬¼´ km•£¨-$\frac{8km}{{4k}^{2}+3}$ £©+m²=0£¬¼´ 3m²=4k²•m²£¬¡àk=$\frac{\sqrt{3}}{2}$£¬Îª¶¨Öµ£®
ÓÉÓÚAB=$\sqrt{{1+k}^{2}}$|x₁-x₂|=$\frac{\sqrt{7}}{2}$•$\sqrt{\frac{24-{4m}^{2}}{3}}$£¬µãOµ½Ö±ÏßABµÄ¾àÀëd=$\frac{|m|}{\sqrt{{1+k}^{2}}}$=$\frac{2|m|}{\sqrt{7}}$£¬
¹ÊS_¡÷AOB=$\frac{1}{2}$•AB•d=$\frac{1}{2}$•$\sqrt{\frac{4£¨6{-m}^{2}£©{•m}^{2}}{3}}$¡Ü$\sqrt{3}$£¬¡à¡÷AOBÃæ»ýµÄ×î´óֵΪ$\sqrt{3}$£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÍÖÔ²µÄ¶¨Òå¡¢±ê×¼·½³Ì£¬ÒÔ¼°¼òµ¥ÐÔÖʵÄÓ¦Ó㬵㵽ֱÏߵľàÀ빫ʽ¡¢ÏÒ³¤¹«Ê½µÄÓ¦Óã¬ÊôÓÚÖеµÌ⣮