过抛物线y^2=12x的焦点作倾斜角为3π/4的弦,则弦长为多少

Question

过抛物线y^2=12x的焦点作倾斜角为3π/4的弦,则弦长为多少

Answer 1

抛物线的焦点为（3,0）,过抛物线的焦点作倾斜角为3π/4的直线
直线的斜率为：k=tan（3π/4）=-1
直线方程为：y-0=k(x-3)=-x+3 即：y=-x+3
直线y=-x+3与抛物线y^2=12x交与A、B两点,设A(x1,y1),B(x2,y2)
y=-x+3 x=3-y代入抛物线方程得：
y^2=36-12y y^2+12y-36=0
y1+y2=-12,y1y2=-36
y1=-x1+3 y2=-x2+3 则有：x1-x2=y2-y1
|AB|=√[(x1-x2)^2+(y1-y2)^2] (将x1-x2=y2-y1代入)
=√[(y1-y2)^2+(y1-y2)^2]
=√2*√[(y1-y2)^2]
=√2*√[(y1+y2)^2-4y1y2] (将y1+y2=-12,y1y2=-36代入)
=24