已知橢圓x^2/a^2+y^2/b^2=1(a>b>0)過點(diǎn)(1,根號(hào)2/2),e=根號(hào)2/2,F1、F2為橢圓有左右焦點(diǎn)P在l:x+y=2上
且y不等于0,直線PF1,PF2與橢圓的交點(diǎn)分別為AB和CD,O為原點(diǎn),(1)求橢圓方程,(2)設(shè)直線PF1、PF2的斜率分別為K1、K2,證明書1/K1-3/K2=2,(3)問直線L上是否存在點(diǎn)P使KOA+KOB+KOC+KOD=0,若存在求出P的所有角解,若不存在說明理由.
答案如下:
![](http://h.hiphotos.baidu.com/zhidao/wh%3D600%2C800/sign=9384a1a894eef01f4d4110c3d0ceb51d/242dd42a2834349b24307649c9ea15ce37d3bec3.jpg)
![](http://e.hiphotos.baidu.com/zhidao/wh%3D600%2C800/sign=32a22af428381f309e4c85af99316030/ac6eddc451da81cb95569b015266d0160824314d.jpg)
![](http://f.hiphotos.baidu.com/zhidao/wh%3D600%2C800/sign=ec89e1436c81800a6eb0810881051fc2/cb8065380cd79123aa0c8cb8ad345982b3b78009.jpg)
此時(shí)直線CD的方程為y=3x1與x+y=2聯(lián)立得x=5/4,y=3/4
![](http://h.hiphotos.baidu.com/zhidao/wh%3D600%2C800/sign=cf661a67033b5bb5be8228f806e3f901/b21c8701a18b87d6c02db0d3070828381e30fd8b.jpg)