Then: d 2 = (x-3) 2+y 2
=(5-3/5*x)^2.
De: 16x 2/25+y 2-25 = 0。 (@)
(1) Let the abscissa of point A be x 1, then it is represented graphically, x1>; 0, so when y=0 in (@), x 1=5.
So: AF=2, (1) is correct.
(2) Let the ordinate of point B be y2, then there is a graph, y2 >;; 0, so when x=0 in (@), y2=4.
So: BF=5, (2) is correct.
(3) by (1), (3) correct.
(4) Errors caused by (2) and (4).
To sum up: the correct one is (1)(2)(3).