过双曲线x^2-y^2/3=1的右焦点F2作倾斜角为π/4的弦AB,求AB的长,

Question

过双曲线x^2-y^2/3=1的右焦点F2作倾斜角为π/4的弦AB,求AB的长,

Answer 1

c=√(a^2+b^2)=√(1+3)=2
双曲线右焦点为F2(2,0)
直线斜率为k=tan(π/4)=1
∴直线方程为y=x-2
代入双曲线,得
x^2-(x-2)^2/3=1
整理得 2x^2+4x-7=0
由韦达定理有：x1+x2=-2,x1x2=-3.5;
y1+y2=(x1-2)+(x2-2)=-2-4=-6,
y1y2=(x1-2)(x2-2)=x1x2-2(x1+x2)+4=-3.5+4+4=4.5
AB=√[(x1-x2)^2+(y1-y2)^2]
=√[(x1+x2)^2-4x1x2+(y1+y2)^2-4y1y2]
=√[(-2)^2-4*(-3.5)+(-6)^2-4*4.5]
=6
∴AB的长为6