Parabola

The point A is the focus point of parabola ${ y }^{ 2 }=4x$ .

Therefore, the line $\overline { PA }$ is same as distance from point P to directrix of parabola ${ y }^{ 2 }=4x$ , which is $x=-1$ .

Let the foot of perpendicular from point P to line $x=-1$ be X. Therefore, the minimum possible distance of $\overline { PA } +\overline { PB }$ is same as $\overline { PX } +\overline { PB }$ .

Euclid postulate states that given any two point, the shortest distance between them is straight line. Therefore, minimum possible value of $\overline { PX } +\overline { PB }$ would be the distance from foot of perpendicular from point B to line $x=1$ to point B.

$\boxed{\therefore \min { \overline { PA } +\overline { PB } } =5+1=6}$

Possible locations of point P are (1,2) or (4,4) calculate the value for PA+PB for these two points and the minimum one will be the answer.

evaluate the value of (x,y) of P as (3,3) ,,, then 1) AP = √(y2-y1)^2+(x2-x1)^2 which equals √(3-0)^2+(3-1)^2 ≈ 3.61 2) BP= √(y2-y1)^2+(x2-x1)^2 which equals √(5-3)^2+(4-3)^2 ≈ 2.24 3) PA+PB ≈ 6

note : if u want to get a closer answer to 6 or even exactly 6 , use all the calculator figures for 1 & 2