The actual curves obtained by cutting a right circular cone by the plane are called conic sections or even conics. Even by varying the inclination of the plane with the axis of the cone several types of curves, viz, the actual parabola, the ellipse and the hyperbola are received. A parabola is really a curve where any point are at an equal length from: a set point (the emphasis ), and a hard and fast straight line (the directrix ). 

A parabola is definitely the locus of a spot which moves to ensure its distance from your focus is equal to its distance from your directrix.
A parabola is usually a symmetrical curve plus it may lie in different orientation in each dimensional plane.

The equation of any parabola opening sideways might be written as y$^2$ = 4ax or maybe y$^2 $ = - 4ax.
Also, situation of parabola beginning upwards or downwards is distributed by x$^2 $ = 4ay or maybe xy$^2 $ = - 4ay.


Parabolic Curve
The following table helps to get the important things about the parabola when the standard equation is known:
Equation
Vertex
 Focus Equation of Directrix
 Axis  Length of Latus Rectum  coordination of ends of latus rectum
 y$^{2}$ = 4ax  (0,0)  (a,0)  x = -a  x axis i.e., y = 0  4a  (a,2a); (a,-2a)
 y$^{2}$ = - 4ax  (0,0)  (-a,0)  x = a  x axis i.e., y = 0  4a  (-a,2a); (-a,-2a)
 x$^{2}$ = 4ay  (0,0)  (0,a)  y = -a  y axis i.e., x= 0  4a  (2a,a); (-2a,a)
 x$^{2}$ = -4ay  (0,0)  (0,-a)  y = a  y axis i.e., x = 0  4a  (2a, -a); (-2a,-a)

The following table can be used to get information from various forms of equations:

Equation
 Vertex  Equation of Axis
Focus  Equation of directrix  Coordinates of ends of lactus rectum
 (y - k)$^{2}$ =4a(x - h)
 (h,k)  y = k  (a+h,k)  x = -a+h  (a+h, 2a+k) and (a+h, -2a+k)
 (y - k)$^{2}$ = -4a(x - h)   (h,k)  y =k  (-a+h,k)  x = a+h  (-a+h, 2a+k) and (-a+h, -2a+k)
 (x - h)$^{2}$ = 4a(y-k)   (h,k)  x = h  (h,a+k)  y = -a+k  (2a+h,a+k) and (-2a+h, a+k)
 (x - h)$^{2}$ = -4a(y - k)   (h,k)  x = h  (h,-a+k)  y = a+k  (2a+h, -a+k) and (-2a+h, -a+k)

The equation of a parabola given within its standard variety, y = ax$^2 $ + bx + h is given under:

Vertex of The Parabola

( - $\frac{b}{2a}$, $\frac{4ac - b^{2}}{4a}$)

Focus of The Parabola

( - $\frac{b}{2a}$, $\frac{4ac - b^{2}+1}{4a}$)

Directrix of The Parabola Formula

y = c - (b$^{2}$ + 1)4a
Some examples on parabola are given below.

Example 1: Find the vertex, equation of the axis, coordinate of the focus, equation of the directrix of the focus, equation of the directrix and coordinates of the end of the lactus rectum of (y-3)$^{2}$ = 6(x-2)

Solution: Comparing with $(y-k)^{2}$ = 4a(x-h)
h = 2, k = 3

Therefore Vertex is (h,k) = (2,3)

Equation of the axis is y = k

i.e., y = 3

4a = 6

a = $\frac{3}{2}$

Focus is (a+h, k) = ($\frac{3}{2}$+ 2, 3)

= ($\frac{7}{2}$,3)

Equation of the directrix is x = -a+h

= -$\frac{3}{2}$+ 2

= $\frac{1}{2}$

i.e., x = $\frac{1}{2}$

Ends of the latus rectum are (a+h, 2a+k)

= ($\frac{3}{2}$ + 2,2.$\frac{3}{2}$  + 3)

= ($\frac{7}{2}$,6)

and (a+h, -2a+k) = ($\frac{3}{2}$ + 2, -2.$\frac{3}{2}$+3)
=($\frac{7}{2}$,0)

Example 2: Find the vertex, equation of the axis, focus, equation of the directrix and coordinates of the end of the latus rectum of $(x-1)^{2}$ = -8(y-2)

Solution: Comparing (x-1)$^{2}$ = -8(y-2) with (x-h)$^{2}$ = -4a(y-k)
h=1, k = 2

Therefore Vertex is (h,k) = (1,2) and 4a = 8
$\therefore$ a = 2

Equation of the axis is x = h
i.e., x = 1

Focus is (h, -a+k)
= (1,-2+2)
=(1,0)

Equation of the directrix is y =a + k
= 2 + 2
i.e., y = 4

Ends of the latus rectum are

(2a+h, -a+k) and (-2a+h, -a+k)

= (2.2+1,-2+2) and (-2.2+1, -2+2)

=(5,0) and (-3,0)

Example 3: Find the equation of the parabola with latus rectum joining the points (-3,1) and (1,1)

Solution: Let L = (-3,1) and L' = (1,1)

The focus S is the midpoint of the latus rectum LL'.

$\therefore$ S = (-1,1)

Since LL' is parallel to x axis, the axis of the parabola is parallel to y axis.

The equation of the parabola is of the form

(x-h)$^{2}$= $\pm$4a(y-k)
......1

Length of the latus rectum = 4a

= $\sqrt{(-3-1)^{2}+ (1-1)^{2}}$

= 4

$\therefore$ a = 1

Case 1: If the parabola open upwards, then
S is (h,a+k)= (-1,1)

$\therefore$  h = -1 and a+ k = 1 since a= 1, k = 0

$\therefore$ (h,k) = (-1,0) is the vertex.

$\therefore$ Equation of the parabola is

(x+1)$^{2}$ = 4(1)(y-0)

i.e., (x+1)$^{2}$ = 4y

Case 2: If the parabola opens downwards, then

S is (h,-a+k) = (-1,1)

$\therefore$ h = -1, -a+k = 1 $\therefore$ k = 2

$\therefore$ Equation of the parabola is (x+1)$^{2}$ = -4(y-2)