Parabola Question Experts please help

Find the vertex, the axis of symmetry, the focus and the directrix of each parabola.

y^2=4(x-1)

The parabola has the form:

(y - k)^2 = 4p(x - h)

or

(x - h)^2 = 4p(y - k)

The parabola y^2 = 4(x - 1) fits the first form above.

The vertex is: (1,0)
The axis of symmetry is: y = 0

4p = 4, so p = 1.

The directrix is: x = 0
The focus is: (2,0)

y^2 = 4(x - 1)
y^2 = 4x - 4
compare to:
(y - k)^2 = 4p (x - h) =>(h , k) is vertex focus is (h + p, k) and the directrix is x = h - p.
vertex at(1 , 0)
the parabola opens right:
axis of symmetry => y = 0 , AKA the x-axis
4p = 4
p = 1
focus at(2 , 0)
directrix=> x = 0

Happy holidays.

x = (1/4)y^2 + 1
a = 1/4, b = 0, c = 1

y-vertex = -b/2a = 0
x-vertex = (1/4)0^2 + 1 = 1
V(1,0)

axis of symmetry is x-axis, equation y = 0

y-focus = -b/2a = 0
x-focus = (1 - b^2 + 4ac)/(4a) = (1 - 0^2 + 4(1/4))/(4 (1/4)) = 2
F(2,0)

directrix
x = - (1 - b^2 + 4ac)/(4a)
x = -2

Vertex (1,0) and 4a = 4 therefore a = 1
axix of symmetry x-1 =0 or x = 1 directrix is x = 1-1 = 0or directrix is x=0

Vertex: (1,0)
Axis of Symmetry: X Axis
Focus: (2,0)
directrix: x = 0