Find the equation of locus of the point which is equidistanc from A(2, 0) and y-axis.

Let P(x, y) be a point on the locus A(2, 0)
Distance from y -axis =|x|
Given PA = |x|
S.O.B.S
$\begin{array}{l}{\mathrm{PA}}^2={\mathrm{x}}^2\\ (\mathrm{x}-2{)}^2+(\mathrm{y}-0{)}^2={\mathrm{x}}^2\\ {\mathrm{x}}^2+4-4\mathrm{x}+{\mathrm{y}}^2-{\mathrm{x}}^2=0;{\mathrm{y}}^2-4\mathrm{x}+4=0\end{array}$
$\text{ The locus of is }{\mathrm{y}}^2-4\mathrm{x}+4=0$