·ÖÎö £¨1£©ÓÉÖ±ÏßÓëÔ²ÏàÇУ¬Çó³öb=1£¬ÓÉ$e=\frac{c}{a}=\frac{{\sqrt{3}}}{2}$£¬µÃ $c=\frac{{\sqrt{3}}}{2}a$£¬ÓÉ´ËÄÜÇó³öÍÖÔ²CµÄ·½³Ì£®
£¨2£©ÓÉÌâÒâ¿ÉÖª£¬Ö±Ïßl2µÄбÂÊ´æÔÚÇÒ²»Îª0£¬¹Ê¿ÉÉèÖ±ÏßlµÄ·½³ÌΪy=kx+m£¨m¡Ù0£©£¬´úÈëÍÖÔ²·½³Ì£¬µÃ£¨1+4k2£©x2+8kmx+4£¨m2-1£©=0£¬ÓÉ´ËÀûÓøùµÄÅбðʽ¡¢Î¤´ï¶¨Àí¡¢µÈ±ÈÊýÁÐÐÔÖÊ¡¢ÏÒ³¤¹«Ê½£¬½áºÏÒÑÖªÌõ¼þ£¬ÄÜÇó³ö¡÷OPQÃæ»ýµÄȡֵ·¶Î§£®
½â´ð ½â£º£¨1£©ÓÉÖ±Ïßl1£º$x-y+\sqrt{2}=0$ÓëÔ²x2+y2=b2ÏàÇУ¬
µÃ£º$d=\frac{{|0+0+\sqrt{2}|}}{{\sqrt{{1^2}+{{£¨-1£©}^2}}}}=1=b$£¬¡£¨2·Ö£©![]()
ÓÉ$e=\frac{c}{a}=\frac{{\sqrt{3}}}{2}$£¬µÃ $c=\frac{{\sqrt{3}}}{2}a$£¬¡£¨3·Ö£©
ÓÖa2=b2+c2£¬¡à${a^2}={1^2}+\frac{3}{4}{a^2}$£¬¡àa2=4£¬¡£¨4·Ö£©
ÍÖÔ²CµÄ·½³ÌΪ$\frac{x^2}{4}+{y^2}=1$¡£¨5·Ö£©
£¨2£©ÓÉÌâÒâ¿ÉÖª£¬Ö±Ïßl2µÄбÂÊ´æÔÚÇÒ²»Îª0£¬¹Ê¿ÉÉèÖ±ÏßlµÄ·½³ÌΪ
y=kx+m£¨m¡Ù0£©£¬P£¨x1£¬y1£©£¬Q£¨x2£¬y2£©£¬
ÓÉ$\left\{\begin{array}{l}{y=kx+m}\\{{x}^{2}+4{y}^{2}-4=0}\end{array}\right.$£¬ÏûÈ¥yµÃ£¨1+4k2£©x2+8kmx+4£¨m2-1£©=0£¬¡£¨6·Ö£©
Ôò¡÷=64k2m2-16£¨1+4k2£©£¨m2-1£©=16£¨4k2-m2+1£©£¾0£¬
ÇÒx1+x2=$\frac{-8km}{1+4{k}^{2}}$£¬x1x2=$\frac{4£¨{m}^{2}-1£©}{1+4{k}^{2}}$£®¡£¨7·Ö£©
¹Êy1y2=£¨kx1+m£©£¨kx2+m£©=k2x1x2+km£¨x1+x2£©+m2£®
¡ßÖ±ÏßOP£¬PQ£¬OQµÄбÂÊÒÀ´Î³ÉµÈ±ÈÊýÁУ¬
¡à$\frac{{y}_{1}}{{x}_{1}}•\frac{{y}_{2}}{{x}_{2}}$=$\frac{{k}^{2}{x}_{1}{x}_{2}+km£¨{x}_{1}+{x}_{2}£©+{m}^{2}}{{x}_{1}{x}_{2}}$=k2£¬¡£¨8·Ö£©
¼´$\frac{-8{k}^{2}{m}^{2}}{1+4{k}^{2}}$+m2=0£¬ÓÖm¡Ù0£¬ËùÒÔk2=$\frac{1}{4}$£¬¼´k=¡À$\frac{1}{2}$£®¡£¨9·Ö£©
ÓÉ¡÷£¾0£¬¼°Ö±ÏßOP£¬OQµÄбÂÊ´æÔÚ£¬µÃ0£¼m2£¼2ÇÒm2¡Ù1£®¡£¨10·Ö£©
S¡÷OPQ=$\frac{1}{2}$|x1-x2||m|=$\sqrt{{m}^{2}£¨2-{m}^{2}£©}$=$\sqrt{1-{{£¨{m^2}-1£©}^2}}$£¬¡£¨11·Ö£©
¡àS¡÷OPQµÄȡֵ·¶Î§Îª£¨0£¬1£©£®¡£¨12·Ö£©
µãÆÀ ±¾Ì⿼²éÍÖÔ²·½³ÌµÄÇ󷨣¬¿¼²éÈý½ÇÐÎÃæ»ýµÄÇ󷨣¬ÊÇÖеµÌ⣬½âÌâʱҪÈÏÕæÉóÌ⣬עÒâ¸ùµÄÅбðʽ¡¢Î¤´ï¶¨Àí¡¢µÈ±ÈÊýÁÐÐÔÖÊ¡¢ÏÒ³¤¹«Ê½µÄºÏÀíÔËÓã®
| 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£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿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£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | £¨0£¬$\frac{1}{3}$£© | B£® | £¨0£¬$\frac{1}{2}$] | C£® | £¨$\frac{1}{3}$£¬$\frac{1}{2}$] | D£® | [$\frac{1}{3}$£¬1£© |
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