#### notes

The equation of a parabola is simplest if the vertex is at the origin and the axis of symmetry is along the x-axis or y-axis. The four possible such orientations of parabola are shown in following fig.

We will derive the equation for the parabola shown above in First Fig with focus at (a, 0) a > 0; and directricx x = – a as below:

Let F be the focus and l the directrix. Let FM be perpendicular to the directrix and bisect FM at the point O. Produce MO to X. By the definition of parabola, the mid-point O is on the parabola and is called the vertex of the parabola. Take O as origin, OX the x-axis and OY perpendicular to it as the y-axis. Let the distance from the directrix to the focus be 2a. Then, the coordinates of the focus are (a, 0), and the equation of the directrix is x + a = 0 as in above fig.

Let P(x, y) be any point on the parabola such that

PF = PB, ... (1)

where PB is perpendicular to l. The coordinates of B are (– a, y).

By the distance formula, we have

PF = `sqrt((x-a)^2+y^2) `and PB = `sqrt(x + a)^2`

Since PF = PB, we have

`sqrt((x-a)^2 + y^2) = sqrt((x+a)^2)`

i.e. `(x – a)^2 + y^2 = (x + a)^2

or x^2 – 2ax + a^2 + y^2 = x^2 + 2ax + a^2`

or `y^2` = 4ax ( a > 0).

Hence, any point on the parabola satisfies

`y^2` = 4ax. ...(2)

Conversely, let P(x, y) satisfy the equation (2)

PF = `sqrt((x-a)^2+y^2) = sqrt((x-a)^2 + 4ax)`

= `sqrt(x+a)^2` = PB .....(3)

and so P(x,y) lies on the parabola.

Standard equation of Parabola:

`y^2` = 4ax

`y^2` = - 4ax

`x^2` = 4ay

`x^2` = - 4ay