Trigonometry help please

Hey, I was having a little trouble with this question and was wondering if someone would be able to help me?

The question says to find an equation in standard form for the parabola that satisfies the given conditions.

Focus (2, -3), Directrix x=5

I'm confused on where to even start this problem, so help is greatly appreciated. Thanks! :)

Focus (2, - 3):

Fx = 2
Fy = - 3

Directrix:

x = 5

The directrix is always perpendicular to the axis of symmetry, so the parabola opens horizontally and, since Fx < x, it opens to the left, so

p = 1/2(Fx - x)
p = 1/2(2 - 5)
p = 1/2(- 3)
p = - 3/2

Vertex (x + p, Fy)
Vertex [5 + (- 3/2), - 3]
Vertex (5 - 3/2, - 3)
Vertex (10/2 - 3/2, - 3)
Vertex (7/2, - 3)

h = 7/2
k = - 3

a = 1 / 4p
a = 1 / 4(- 3/2)
a = 1 / (- 12/2)
a = 1 / - 6
a = - 1/6

x = a(y - k)² + h
x = - 1/6[y - (- 3)]² + 7/2
x = - 1/6(y + 3)² + 7/2
x = - 1/6(y² + 6y + 9) + 7/2
x = - 1/6 y² - y - 9/6 + 21/6
x = - 1/6 y² - y + 12/6
x = - 1/6 y² - y + 2
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Since the directrix is a vertical line x=5, the parabola opens right (right/left depends on whether the line is to the right or left of the y-axis, i.e., x=p with p>0 or x=p with p<0). Here p=5

The equation then x=(1/4p) y^2 =y^2/20

The focus is p units to the right of the y-axis at (5,0)