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

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$(y-2)^2=4(x-1)$------(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.

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