$\text{ Consider the parabola }{y}^2=4x\text{. Let }A=(-1,0)\text{ and }B=(0,1).F\text{ is the focus of the parabola}$

$\text{ If }P(\alpha ,\beta )\text{ is a point on the parabola such that |}PA-PB|\text{is maximum then α+β=}$

$|\mathrm{PA}-\mathrm{PB}|\text{ is max when }\mathrm{P},\underset{\_}{\mathrm{A},\mathrm{B}}\text{ are collinear and }P\text{ divides }\mathrm{AB}\text{ externally }$

$\begin{array}{l}\text{ Equation of }AB\text{ is }-x+y=1\text{. i.e., }y=x+1\\ (x+1{)}^2=4x\Rightarrow x=1\end{array}$

$\overrightarrow{\mathrm{AB}}\text{ intersect parabola at }(1,2)$

$The sum of the coordiantes is 3 units$