Finding standard form of parabola equation I have a little problem with figuring it out how find standard form of a parabola equation from the given values. I tried googling and watching video in youtube but I don't understand how to actually go about it.
Given:

*

*Directrix is $ x = -2$ ; focus is $(2,0)$

*Axis along the $x$ axis; vertex at the origin; passing through $(2,5)$

*Vertex at $(-2,-2)$; focus at $(-2,2)$
I don't know how to formulate this, some hints would be appreciated.
Edit. I read this but I don't understand.
 A: Note that it is important to know the definition of a parabola, what the curve signifies besides the geometry. A parabola is a curve (equation) where any point is equidistant from a fixed point (focus) and a fixed line (directrix). Have a look at the image below to get familiar with the terms, first.
Your question seems to describe 3 different parabolas.
Parabola 1. Directrix $\equiv x=-2$ , focus $\equiv (2,0)$
We go with the definition. Let $(h,k)$ be an arbitrary point on the parabola. Then, it must be equidistant from the directrix, $x+2=0$ and focus, $(2,0)$ i.e., it satisfies
\begin{align}|h+2|=\sqrt{(h-2)^2+k^2}&\Rightarrow (h+2)^2=(h-2)^2+k^2\\
&\Rightarrow k^2=(h+2)^2-(h-2)^2=8h\\
&\Rightarrow k^2=8h.\end{align}
Since, $(h,k)$ was an arbitrary point on the curve, we can say that the equation of the parabola is nothing but
$$y^2=8x.$$
Parabola 2. Axis $\equiv y=0$ ($x$ axis) , vertex $\equiv (0,0)$ , $(2,5)$ is a point lying on the curve
The ''axis of symmetry'' or simply, the axis of a parabola is the line perpendicular to the directrix and passing through the focus of the parabola. The vertex has been defined as the point where the axis meets the curve itself.
Any parabola with $y=0$ as its axis of symmetry and vertex $(0,0)$ is of the form
$$y^2=4px\mathrm{~where~}(p,0)\mathrm{~is~the~focus}.$$

(Why is this?)

The focus must be a point lying on the $x$ axis and hence, of the form $(p,0)$. Since, the vertex is a point on the parabola, it must be equidistant from the directrix and the focus. The axis is perpendicular to the directrix, passing through the focus and vertex. Suppose the axis meets the directrix at point $D$ and let $F$ be the focus. Then, the vertex, $(0,0)$ is the midpoint of the line segment, $FD$ and $D$ lies on the axis. We get,
$$D\equiv (-p,0).$$
We use this to get the equation of the parabola, as we did in part 1:
\begin{align}|x+p|=\sqrt{(x-p)^2+y^2}&\Rightarrow (x+p)^2=(x-p)^2+y^2\\
&\Rightarrow y^2=(x+p)^2-(x-p)^2=4px\\
&\Rightarrow y^2=4px.\end{align}

We know that $(2,5)$ also lies on the parabola. Therefore, it satisfies the equation $y^2=4px$ i.e.,
$$5^2=4p\cdot2\Rightarrow p=25/8.$$
Equation of the parabola becomes
$$y^2=\frac{25}{2}x\Leftrightarrow 2y^2=25x.$$
Parabola 3. Vertex $\equiv (-2,-2)$, focus $\equiv (-2,2)$
The focus and vertex of a parabola lie on the axis and, the vertex is the midpoint of the line segment joining the focus and the point where the axis meets the directrix (let us call this point $D$). The directrix will be a line perpendicular to the axis and passing through $D$.
First we find $D\equiv(h,k)$. Then,
$$\frac{h-2}{2}=-2\Rightarrow h=-2\mathrm{~and~}\frac{k+2}{2}=-2\Rightarrow k=-6.$$
$\therefore D\equiv(-2,-6).$
Next, we observe that the line $x=-2$ passes through $(-2,-2)$ and $(-2,2)$. Since, any two points in $\mathbb{R}^2$ have exactly one straight line passing through both of them, we can conclude that the equation of the axis is
$$x+2=0.$$

Any line perpendicular to some line, $ax+by+c=0$ is of the form $bx-ay+k=0.$

The directrix is of the form
$$y-k=0$$
and it passes through the point $D\equiv(-2,-6).$ Hence, $k=-6$ and the directrix is $y+6=0$. Using these, we get that the equation of the parabola is
\begin{align}|y+6|=\sqrt{(x+2)^2+(y-2)^2}&\Rightarrow (y+6)^2=(x+2)^2+(y-2)^2\\
&\Rightarrow (x+2)^2=(y+6)^2-(y-2)^2=8(2y+4)\\
&\Rightarrow (x+2)^2=16(y+2).\end{align}

Note. Once you get familiar with parabolas, you'll learn some standard parabolas and their properties. You can then solve for parabola 3 using @Jan-Magnus Økland's comment under your question.

