1.
设A(x1,y1),B(x2,y2),左焦点(-c,0)
则直线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
根据焦半径公式有
|AF1|=a+ex1
|BF1|=a+ex2
∴|AB|=|AF1|+|BF1|=2a+e(x1+x2)=4a/3
∴e(x1+x2)=-2a/3
联立椭圆和直线
y=x+c
x²/a² + y²/b² =1,得
(a²+b²)x²+2a²c+a²c²-a²b²=0
把b²=a²-c²代入,得
(2a²-c²)x²+2a²cx+(2c²-a²)a²=0
∴e(x1+x2)=e[-2a²c/(2a²-c²)]=-2a/3
e(ac)/(2a²-c²)=1/3 (左右约去-2a)
e(c/a)/[2-(c/a)²]=1/3 (上下同时除以a²)
e²/(2-e²)=1/3
e=√2/2
2.
PA=PB
即(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 (把y=x+c代入)
(x1-x2)(x1+x2+2) + (x1-x2)(x1+x2+2c)=0
(x1-x2)[2(x1+x2)+2+2c]=0
∵x1≠x2,即x1-x2≠0
∴2(x1+x2)+2+2c=0
∴x1+x2+1+c=0
即
[-2a²c/(2a²-c²)]+1+c=0
∵e=c/a=√2/2,即a²=2c²
代入上式,得
c=3
∴a=3√2,a²=18,b²=9
椭圆方程为x²/18+y²/9=1
