Standard Equations of a Parabola
To obtain the equation of a parabola in a simplified form, we take the origin as the vertex V , the x axis as axis of symmetry . Let us take VF = a, so that the coordinates of F are (a, 0). Since the directrix D is at a distance a from the vertex and is parallel to yaxis, its equation is x + a = 0.
Let P (x, y) be any point on the parabola. Let M be the foot of perpendicular from P to the directrix D. By definition PF = PM.
We have PF =
and PM = x + a
Since PF = PM, we get PF^{2} = PM^{2}â‡’ (x  a)^{2} + y^{2} = (x + a)^{ 2 }
â‡’ y^{2} = (x + a)^{ 2}  (x  a)^{ 2} = 4ax â‡’ y^{2} = 4ax.
Shape of Parabolic Curve: y2 = 4ax (a > 0)
 x cannot be negative, because for x < 0, y is imaginary. Thus, no part of parabola lies to the left of yaxis. That is, the entire curve lies to the right side of yaxis.
 The curve passes through the origin, since the origin (0, 0) satisfies its equation. Further, x = 0 â‡’ y^{2 }= 0, so the yaxis touches the curve at (0, 0).
 From y^{2 }= 4ax we get y = Â± 2(ax)^{1/2}. This means, corresponding to one positive value of x, there are two, equal and opposite values of y. This shows that the parabolic curve is symmetrical about the xaxis.
 As x increases, y also increases and there is no limit to this joint increase of x and y. That is, as x â†’ âˆž , y â†’ Â± âˆž.
In the following table, we list the graph of four parabolas. Each of these parabolas has its vertex at the origin and the focus either on xaxis or yaxis. We shall take a > 0.
Vertex 
Focus 
Directrix 
Equation 
Description 
Graph 
(0,0) 
(a, 0) 
x = a 
y^{2 }= 4ax 
Parabola, axis of symmetry is the xaxis, opens to the right 

(0,0) 
(a,0) 
x = a 
y^{2 }= 4ax 
Parabola, axis of symmetry is the x  axis, opens to the left 

(0, 0) 
(0, a) 
y = a 
x^{2 }= 4ay 
Parabola, axis of symmetry is the yaxis, opens up 

(0, 0) 
(0, a) 
y = a 
x^{2 }= 4ay 
Parabola, axis of symmetry is the yaxis, opens down 