$$x^2 + 6x \color{blue}{+ 9} + 8y + 1 \color{blue}{- 9} = 0$$
$$(x + 3)^2 + 8(y - 1) = 0$$
$$(x + 3)^2 = -8(y - 1) = 4(-2)(y-1)$$
Now, you should be able to "read off" the vertex of the parabola. From there, see if you can find.
With respect to completing the square: you have
$$(x + 3)^2 + 8 y + 1 = 9$$
Subtract $9$ from both sides of the equation. $$\begin{align} (x + 3)^2 + 8y + 1 - 9 = 0 & \iff (x+3)^2 + 8y - 8 = 0 \\ \\ & \iff (x+3)^2 + 8(y - 1) = 0 \end{align}$$
Then subtract $8(y - 1)$ from each side of the equation to get the form you need:
$$(x + 3)^2 = -8(y - 1) = 4(-2)(y - 1)$$
$$(x + 3)^2 = 4(-2)(y - 1)$$
So, $$p = -2, \;h = -3,\; k = 1$$
Here, it is not equation of axis, it is perpendicular distance of any random point on parabola say (h,k) from axis and also perpendicular distance from vertex and not equation of vertex
So for general equation
Perpendicular distance from axis of point (h,k) is k and from vertex, it's h
So equation will be:
k²=4ah
Replacing h with x and k with y, we get equation of parabola
y²=4ax
Best Answer
Using some trigonometry:
Let $ y = a(x-h)^2 + k $ represent the equation of the parabola at vertex $(h, k)$ on the $xy$ plane
A two dimensional rotation is described as $$x = x'\cos(\theta) + y'\sin(\theta)$$ $$y = -x'\sin(\theta) + y'\cos(\theta)$$ where $(x', y')$ is a point on the $x'y'$plane rotated by $\theta$
Substituting these formulas into the equation of the parabola we get
$$-x'\sin\theta+y'\cos(\theta)=a(x'\cos(\theta) + y'\sin(\theta) - h)^2+k$$
which is probably the most useful form as simplifying gets messy
Here is a desmos link with a rotating parabola: https://www.desmos.com/calculator/hguanwbkbu