Let ABC be three different points on the parabola y2 = 4x, if the focus F of the parabola is exactly the center of gravity It is known that ABC is three different points on the parabola y ^ 2 = 4x. If the focus F of the parabola is exactly the center of gravity of the triangle ABC, then the value of AF + BF + CF is equal to?

Let ABC be three different points on the parabola y2 = 4x, if the focus F of the parabola is exactly the center of gravity It is known that ABC is three different points on the parabola y ^ 2 = 4x. If the focus F of the parabola is exactly the center of gravity of the triangle ABC, then the value of AF + BF + CF is equal to?

This paper mainly studies the definition of parabola
Focus f (1,0), because f is the center of gravity, so the sum of abscissa of ABC three points should be 3 times of abscissa of F point, that is 3, and the distance AF from a to focus should be equal to the distance from a to the collimator x = - 1, that is x (a) + 1, similarly BF = x (b) + 1, CF = x (c) + 1, so AF + BF + CF = 3 + 3 = 6