Question

椭圆X2/4 y2/3=1的左右焦点分别为F1F2,过F1直线交椭圆于AB两点已知△ABF1面积为12厂2/7求AB直线

Answer 1

c=1,F1(-1,0),F2(1,0),

设AB:x=my-1,

代入x^2/4+y^2/3=1得3(m^2y^2-2my+1)+4y^2=12,

整理得(3m^2+4)y^2-6my-9=0,

△=36m^2+36(3m^2+4)=144(m^2+1),

|AB|=√[△(1+m^2)]/(3m^2+4)=12(m^2+1)/(3m^2+4),

F2到AB的距离h=2/√(m^2+1),

∴S△ABF2=(1/2)|AB|h=12√(m^2+1)/(3m^2+4)=12√2/7,

∴7√(m^2+1)=(3m^2+4)√2,

平方得49(m^2+1)=2(9m^4+24m^2+16),

整理得18m^4-m^2-17=0,m^2>=0,

∴m^2=1,m=土1,

∴AB的方程是x土y+1=0.