·ÖÎö £¨1£©£ºÓÉÌâÒâ¿ÉµÃ£ºe=$\frac{\sqrt{2}}{2}$=$\frac{c}{a}$£¬2a+2c=4+2$\sqrt{2}$£¬ÓÖa2=b2+c2£®ÁªÁ¢½â³ö¼´¿ÉµÃ³öÍÖÔ²CµÄ·½³Ì£®
£¨2£©ÉèD£¨x0£¬y0£©£¬Ôò$\frac{{x}_{0}^{2}}{4}$+$\frac{{y}_{0}^{2}}{2}$=1£®°Ñy=m´úÈëÍÖÔ²·½³Ì¿ÉµÃ£ºA£¨-$\sqrt{4-2{m}^{2}}$£¬m£©£¬B£¨$\sqrt{4-2{m}^{2}}$£¬m£©£®ÀûÓõãбʽ¿ÉµÃ£ºÖ±ÏßDAµÄ·½³ÌÓëÖ±ÏßDBµÄ·½³Ì£¬¿ÉµÃP£¬QµÄ×ø±ê£®ÀûÓÃбÂʹ«Ê½Ö»ÒªÖ¤Ã÷${k}_{P{F}_{1}}$•${k}_{Q{F}_{1}}$=1¼´¿ÉµÃ³ö£®
½â´ð £¨1£©½â£ºÓÉÌâÒâ¿ÉµÃ£ºe=$\frac{\sqrt{2}}{2}$=$\frac{c}{a}$£¬2a+2c=4+2$\sqrt{2}$£¬ÓÖa2=b2+c2£®
ÁªÁ¢½âµÃ£ºa=2£¬b=c=$\sqrt{2}$£®
¡àÍÖÔ²CµÄ·½³ÌΪ£º$\frac{{x}^{2}}{4}+\frac{{y}^{2}}{2}$=1£®
£¨2£©½â£º¡ÏPF1F2+¡ÏQF1F2=90¡ã£®
ÏÂÃæ¸ø³öÖ¤Ã÷£ºF1$£¨-\sqrt{2}£¬0£©$£®
ÉèD£¨x0£¬y0£©£¬Ôò$\frac{{x}_{0}^{2}}{4}$+$\frac{{y}_{0}^{2}}{2}$=1£®
°Ñy=m´úÈëÍÖÔ²·½³Ì¿ÉµÃ£º$\frac{{x}^{2}}{4}$+$\frac{{m}^{2}}{2}$=1£¬½âµÃx=¡À$\sqrt{4-2{m}^{2}}$£®
È¡A£¨-$\sqrt{4-2{m}^{2}}$£¬m£©£¬B£¨$\sqrt{4-2{m}^{2}}$£¬m£©£®
Ö±ÏßDAµÄ·½³ÌΪ£ºy-y0=$\frac{m-{y}_{0}}{-\sqrt{4-2{m}^{2}}-{x}_{0}}$£¨x-x0£©£¬¿ÉµÃP$£¨0£¬\frac{£¨m-{y}_{0}£©{x}_{0}}{\sqrt{4-2{m}^{2}}+{x}_{0}}+{y}_{0}£©$£®
ͬÀí¿ÉµÃ£ºÖ±ÏßDBµÄ·½³ÌΪ£ºy-y0=$\frac{m-{y}_{0}}{\sqrt{4-2{m}^{2}}-{x}_{0}}$£¨x-x0£©£¬¿ÉµÃQ$£¨0£¬\frac{-{x}_{0}£¨m-{y}_{0}£©}{\sqrt{4-2{m}^{2}}-{x}_{0}}+{y}_{0}£©$£®
¡à${k}_{P{F}_{1}}$=$\frac{m{x}_{0}+{y}_{0}\sqrt{4-2{m}^{2}}}{\sqrt{2}£¨\sqrt{4-2{m}^{2}}+{x}_{0}£©}$£¬
${k}_{Q{F}_{1}}$=$\frac{-m{x}_{0}+{y}_{0}\sqrt{4-2{m}^{2}}}{\sqrt{2}£¨\sqrt{4-2{m}^{2}}-{x}_{0}£©}$£®
ÓÖ${y}_{0}^{2}$=2-$\frac{{x}_{0}^{2}}{2}$£®
¡à${k}_{P{F}_{1}}$•${k}_{Q{F}_{1}}$=$\frac{m{x}_{0}+{y}_{0}\sqrt{4-2{m}^{2}}}{\sqrt{2}£¨\sqrt{4-2{m}^{2}}+{x}_{0}£©}$•$\frac{-m{x}_{0}+{y}_{0}\sqrt{4-2{m}^{2}}}{\sqrt{2}£¨\sqrt{4-2{m}^{2}}-{x}_{0}£©}$=$\frac{{y}_{0}^{2}£¨4-2{m}^{2}£©-{m}^{2}{x}_{0}^{2}}{2£¨4-2{m}^{2}-{x}_{0}^{2}£©}$=$\frac{£¨2-\frac{{x}_{0}^{2}}{2}£©£¨4-2{m}^{2}£©-{m}^{2}{x}_{0}^{2}}{2£¨4-2{m}^{2}-{x}_{0}^{2}£©}$=1£®
¡à¡ÏPF1F2+¡ÏQF1F2=90¡ã£®
µãÆÀ ±¾Ì⿼²éÁËÍÖÔ²µÄ±ê×¼·½³Ì¼°ÆäÐÔÖÊ¡¢Ö±ÏßÓëÍÖÔ²ÏཻÎÊÌâ¡¢Ö±Ïß·½³Ì¡¢Ð±ÂʼÆË㹫ʽ¡¢µãÓëÍÖÔ²µÄλÖùØÏµ£¬¿¼²éÁË̽¾¿ÄÜÁ¦¡¢ÍÆÀíÄÜÁ¦Óë¼ÆËãÄÜÁ¦£¬ÊôÓÚÄÑÌ⣮
| Äê¼¶ | ¸ßÖÐ¿Î³Ì | Äê¼¶ | ³õÖÐ¿Î³Ì |
| ¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | 30¡ã | B£® | 45¡ã | C£® | 60¡ã | D£® | 75¡ã |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | {x|-3£¼x£¼-1} | B£® | {x|-3£¼x£¼0} | C£® | {x|-1¡Üx£¼0} | D£® | {x|x£¼-3} |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¹ú¼ÊѧУÓÅÑ¡ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com