设抛物线y^2=8x的焦点为F,有倾斜角为45°的直线交抛物线于A,B两点,AB的距离为8√5,求△FAB的面积

y^2=8x的焦点为F(2,0)有倾斜角为45°的直线L交抛物线于A,B两点L:y=x+b,x=y-by^2=8x=8(y-b)y^2-8y+8b=0yA+yB=8,yA*yB=8b(xA-xB)^2=(yA-yB)^2=(yA+yB)^2-4yA*yB=8^2-4*8b=64-32bAB^2=(xA-xB)^2+(yA-yB)^2=2*(64-32b)AB...