Question

椭圆C的焦点在x轴上,焦距为2,直线n:x-y-1=0与椭圆C交于A、B两点,F1是左焦点,且F1A┴F1B,则椭圆C的

Answer 1

设椭圆方程标准方程为：(x²/a²)+(y²/b²)=1（a＞b＞0）

已知2c=2,所以c=1

则,a²=b²+1

即,x²/(b²+1)+y²/b²=1

===> b²x²+(b²+1)y²-b²(b²+1)=0

联立直线与椭圆方程得到：b²x²+(b²+1)(x-1)²-b²(b²+1)=0

===> (2b²+1)x²-2(b²+1)x+(b²+1)(1-b²)=0

===> (2b²+1)x²-2(b²+1)x+(1-b^4)=0

===> x1+x2=2(b²+1)/(2b²+1)；x1x2=(1-b^4)/(2b²+1)

设A(x1,x1-1)；B(x2,x2-1)

已知F1(-1,0)

因为F1A⊥F1B,则：Kf1a*Kf1b=-1

===> [(x1-1)/(x1+1)]*[(x2-1)/(x2+1)]=-1

===> (x1-1)(x2-1)=-(x1+1)(x2+1)

===> x1x2-(x1+x2)+1=-x1x2-(x1+x2)-1

===> x1x2=-1

===> (1-b^4)/(2b²+1)=-1

===> 1-b^4=-2b²-1

===> b^4-2b²-2=0

===> (b²-1)=3

===> b²-1=±√3

===> b²=√3+1,或者b²=-√3+1＜0,舍去

那么,a²=b²+1=√3+2

所以,椭圆的标准方程为：x²/(√3+2)+y²/(√3+1)=1