设F1,F2,分别是椭圆C：(x^2/a^2)+(y^2/b^2)=1(a>b>0)的左,右焦点,过F1且斜率为1的直线

1.设A(x1,y1),B(x2,y2),左焦点(-c,0),直线l：y=x+c
由题意得|AF2|+|BF2|=2|AB|
∵ |AF1|+|AF2|=2a. .①
|BF1|+|BF2|=2a.②
①+②得(|AF1|+|BF1|)+(|AF2|+|BF2|)=4a
即|AB|+2|AB|=4a,|AB|=4a/3……③
直线y=x+c代入椭圆方程x²/a² + y²/b² =1,得(a²+b²)x²+2a²cx+a²c²-a²b²=0
x1+x2=-2a²c/(a²+b²),x1*x2=(a²c²-a²b²)/(a²+b²)
|AB|=√(1+k²)|x1-x2|=√2*√[(x1+x2)²-4x1x2]
把c²=a²-b²代入,得|AB|=4ab²/(a²+b²)……④
由③④得a²=2b²,a=√2b
（2）设点P(0,1)满足|PA|=|PB|,则|PA|²=|PB|²
所以(x1+1)²+y1²=(x2+1)²+y2²
(x1+1)²-(x2+1)²+y1²-y2²=0
(x1-x2)(x1+x2+2) + (y1-y2)(y1+y2)=0
(x1-x2)(x1+x2+2) + [(x1+c)-(x2+c)][(x1+c)+(x2+c)]=0
(x1-x2)(x1+x2+2) + (x1-x2)(x1+x2+2c)=0
2(x1-x2)[(x1+x2)+1+c]=0
∵x1≠x2,∴x1-x2≠0,∴(x1+x2)+1+c=0
即-2a²c/(a²+b²)+1+c=0
∵a²=2b²,c²=a²-b²,得到b=c
代入上式整理得b=3,a=18
椭圆C方程x²/18+y²/9=1