过抛物线y^2=4x的焦点作倾斜角为π/3的弦AB,求AB

Question

过抛物线y^2=4x的焦点作倾斜角为π/3的弦AB,求AB

Answer 1

由y^2=2px=4x (1)得2p=4,p=2.
抛物线的焦点F(p/2,0) ---->F(1,0)
过焦点F且倾斜角为π/3的直线的方程为：y=k(x-1)=tan(π/3)(x-1).
故 y=√3(x-1).(2)
(1)、(2)式,得：
x1=3,x2=1/3.
y1=2√3.y2=-2√3/3.
即,A(3,2√3),B(1/3,-2√3/3)
|AB|=√{[3-(1/3)]^2+[2√3-(-2√3/3)]^2}
=16/3 .
所求的弦AB的长度为16/3 (长度单位）.