花花m9yg2b
幼苗
共回答了22个问题采纳率:86.4% 举报
(1)①易知
![](https://img.yulucn.com/upload/0/d9/0d946f02c6e043809b05f74853d4d3a0_thumb.jpg)
,
∴b
2 =3,
又F(1,0),∴c=1,a
2 =b
2 +c
2 =4
∴椭圆C的方程为
②∵l与y轴交于M
设A(x
1 ,y
1 ),B(x
2 ,y
2 ),由
∴(3m
2 +4)y
2 +6my﹣9=0,△=144(m
2 +1)>0 ∴
又由
![](https://img.yulucn.com/upload/2/43/243db919261d40b0d36b788759950a5a_thumb.jpg)
,
∴
![](https://img.yulucn.com/upload/c/7e/c7eabebc971048df87cce21d73ce709f_thumb.jpg)
∴
同理
∴
(2)∵F(1,0),k=(a
2 ,0),
先探索,当m=0时,直线l ⊥Ox轴,则ABED由对称性知,AE与BD相交FK中点N,且
猜想:当m变化时,AE与BD相交于定点
证明:设A(x
1 ,y
1 ),B(x
2 ,y
2 ),E(a
2 ,y
2 ),D(a
2 ,y
1 )
当m变化时首先AE过定点N
∵
![](https://img.yulucn.com/upload/3/54/35498a8ab0385a6112690904cd23d8b8_thumb.jpg)
,即(a
2 +b
2 m
2 )y
2 +2mb
2 y+b
2 (1﹣a
2 )=0
又△=4a
2 b
2 (a
2 +m
2 b
2 ﹣1)>0(a>1)
又K
AN =
![](https://img.yulucn.com/upload/f/1e/f1e628a88ea0be2276a9e62b59df71b0_thumb.jpg)
,
而K
AN ﹣K
EN =
![](https://img.yulucn.com/upload/6/db/6dbf73abb69dd2fa774672cf1d536d1b_thumb.jpg)
=
∴K
AN =K
EN ,∴A、N、E三点共线,
同理可得B、N、D三点共线∴AE与BD相交于定点
1年前
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