Find the vertex,lactus rectum,axis and the directix of the parabola $(y-2)^2=4(x-1)$?

$\begin{array}{1 1}(a)\;(1,5),y=3,x=1\\(b)\;(1,3),y=3,x=1\\(c)\;(1,2),y=2,x=0\\(d)\;(2,2),y=4,x=2\end{array}$

$(y-2)^2=4(x-1)$------(1)
Let make the parabola in form of $y^2=4ax$
Hence making
$Y=y-2$
$X=x-1$
Hence replacing makes equ(1)
$Y^2=4aX$
Now vertex for $Y^2=4aX$ is (0,0)
Hence $X=0,Y=0$
$y-2=0\Rightarrow y=2$
$x-1=0\Rightarrow x=1$
(1,2) are the vertex of $(y-2)^2=4(x-1)$
Focus (a,0)
$a=1$
Focus (1,0)
Axis for $Y^2=4aX$ is $Y=0$
Hence $y=2$ is the axis of the parabola $(y-2)^2=4(x-1)$
Equation of directix for $Y^2=4aX$ is $X+a=0$
For $(y-2)^2=4(x-1)$ is
$x-1+1=0$
$x=0$
Hence (c) is the correct answer.