Show that the equation of a parabola in the standard form is y² = 4ax

Let S be the focus and L = 0 be the directrix of the parabola.

 Let S be focus on the right side of L = 0. 

Let P be a point on the parabola.

 Let M and Z be the projections (feet of the perpendiculars) of P and S respectively on the directrix.
Let N be the projection of P on $\overleftrightarrow{\mathrm{SZ} }$. 

Let A be the mid point of $\overleftrightarrow{\mathrm{SZ} }$. 

Then SA = AZ and hence, by definition, A lies on the parabola (A is called the vertex of the parabola).
Let us consider  $\overrightarrow{\mathrm{AS}}$ the principal axis as the positive X-axis and $\overrightarrow{\mathrm{AY}}$ perpendicular to $\overrightarrow{\mathrm{AS}}$,as the positive y–axis. Then A = (0, 0), the origin.
Let AS=a then S = (a, 0), Z = (–a, 0) and the equation of the directrix is x + a = 0
Let P = (x₁, y₁) 

Since P lies on the parabola then
$\begin{array}{l}\Rightarrow \frac{\mathrm{SP}}{\mathrm{PM}}=1\\ \Rightarrow \mathrm{SP}=\mathrm{PM}\end{array}$
From the diagram $\mathrm{PM}=\mathrm{NZ}=\mathrm{NA}+\mathrm{AZ}={\mathrm{x}}_1+\mathrm{a}$
Now,
$\begin{array}{l}\mathrm{SP}=\mathrm{PM}\Rightarrow \sqrt{{\left({\mathrm{x}}_1-\mathrm{a}\right)}^2+{\mathrm{y}}_1^2}=\left({\mathrm{x}}_1+\mathrm{a}\right)\\ \Rightarrow {\left({\mathrm{x}}_1-\mathrm{a}\right)}^2+{\mathrm{y}}_1^2={\left({\mathrm{x}}_1+\mathrm{a}\right)}^2\\ \Rightarrow {\mathrm{y}}_1^2={\left({\mathrm{x}}_1+\mathrm{a}\right)}^2-{\left({\mathrm{x}}_1-\mathrm{a}\right)}^2\\ \Rightarrow {\mathrm{y}}_1^2=4{\mathrm{ax}}_1\end{array}$
The locus of P is y² = 4ax.