抛物线C的方程为y2=4x,焦点为F,准线与x轴的交点为K.过点F作倾斜角为兀/4的直线交抛物线C于A,B两点,则三角形ABK的面_数学解答

抛物线C的方程为y2=4x,焦点为F,准线与x轴的交点为K.过点F作倾斜角为兀/4的直线交抛物线C于A,B两点,则三角形ABK的面积为

人气：461 ℃　时间：2020-06-21 22:25:59

解答

F(1,0) K(-1,0)
直线：y=x-1
(x-1)^2=4x
x^2-6x+1=0
x1+x2=6 x1x2=1
y1+y2=4 y1y2=(x1-1)(x2-1)=x1x2-(x1+x2)+1=1-6+1=-4
S△ABK=S△FBK+S△FAK
=1/2*|FK|*|y1|+1/2*|FK|*|y2|
=|y1|+|y2|
=|y1-y2|
=根号(y1-y2)^2
=根号[(y1+y2)^2-4y1y2]
=根号(4^2+4*4)
=4根号2
=