Take the fixed point m (4,0) as a straight line L, intersect with the parabola y2=4x at two points A and B, and f is the focus of the parabola, so as to find the minimum value of the triangle AFB area

Take the fixed point m (4,0) as a straight line L, intersect with the parabola y2=4x at two points A and B, and f is the focus of the parabola, so as to find the minimum value of the triangle AFB area. Let the linear l equation be x-4=my, and substitute y? =4x, get

y？ =4my+ 16

y 1+y2=4m

y 1*y2=- 16

△ the area of △AFB is s = (4-1) | y1-y2 |/2 = 3 √ [(y1+y2)? -4y 1*y2] /2=6√(m？ +4)

When m=0, the minimum area is 12.