If the sum of the distances from a point a to B (3,2) on the parabola y2 = 4x to the focus is the smallest, then the coordinate of point a is______ ．

If the sum of the distances from a point a to B (3,2) on the parabola y2 = 4x to the focus is the smallest, then the coordinate of point a is______ ．

From the parabola y2 = 4x, we can get the focus f (1,0) and the equation of the straight line L: x = - 1. As shown in the figure, when passing through point a, let am ⊥ L and the perpendicular foot be m, then | am | = | AF |. Therefore, when three points B, a and m are collinear, the minimum value of | ab | + | am | = | BM | is 3 - (- 1) = 4

Given the point P (x, y) and point a (a, 0) (a ∈ R) on the parabola y square = 2x, let the minimum distance from P to a be f (a)
Given the point P (x, y) and point a (a, 0) (a ∈ R) on the parabola y square = 2x, let the minimum distance from P to a be f (a). (1) find the expression of F (a) (2) when 1 / 3 ≤ a ≤ 5, find the maximum and minimum of F (a)

（1） Let P (2t & sup2;, 2T, and (t ∈ R) be set point P (2t & sup2; (2t & sup2; (2t & sup2; (2t & sup2; (2t & sup2; - a) & sup2; (4T & sup2; + 4T & sup2; = [2T & sup2; (A-1; (A-1)] & sup2; (2t & sup2;, 2T & sup2; (2t (2t & sup2; (2t & sup2; (2t & sup2; (2t-2; (a-a-2; (a-1-1-1)] (P (p-pa | & sup2 | & sup2; (a | & sup2; (a \\\124124; & sup2; (a; (f (a) = √ (2a-1), (a ＞ 1)]. (2) when a ∈ [1 / 3,5], the combination of numbers and shapes shows that, f(a)min=f(1/3)=1/3,f(a)max=f(5)=3.

The shortest distance from point P (x, y) on parabola y ^ 2 = 2x to point a (a, 0) (a ∈ R) is f (a). Find f (a)
(1) Find f (a)
(2) When 1 / 3 ≤ a ≤ 5, find the maximum and minimum of F (a)

(1) F (a) = | PA | = √ [(x-a) ^ 2 + y ^ 2] ∵ P on the parabola  f (a) = √ [(x-a) ^ 2 + 2x = √ (x ^ 2 + A ^ 2-2ax + 2x) (2) f '(a) = 1 / 2 (x ^ 2 + A ^ 2-2ax + 2x) ^ (- 1 / 2) (A-X) Let f' (a) = 0 deduce a = x, then f '(a) = √ 2xf' (a) max = √ 10F '(a) min = √ (2 / 3)

Find a point P on the parabola y ^ 2 = 2x, so that the sum of the distances from P to focus F and to point a (3,2) is the smallest

Obviously f is (1 / 2,0)
Let the distance from the point P of the parabola to the directrix be | PQ|
According to the definition of parabola:
|PF|=|PQ|
∴|PF|+|PA|=|PQ|+|PA|
When p, Q and a are collinear
|PQ | + | PA | minimum
∵ a (3,2), let P (x1,2) be substituted into y ^ 2 = 2x to get: X1 = 2
So the coordinates of point P are (2,2)