Question

A parabola can be drawn given a focus of (0, 2) and a directrix of y = 0. Write the

equation of the parabola in any form.

Answer 1

Given:

Focus of a parabola = (0,2)

Directrix: y=0.

To find:

The equation of parabola.

Solution:

The equation of parabola is

`y=(1)/(4p)(x-h)^2+k` ...(i)

where, (h,k) is vertex, (h,k+p) is focus, y=k-p is directrix.

Focus :
`(h,k+p)=(0,2)`

On comparing both sides, we get

`h=0`

`k+p=2` ...(ii)

On comparing y=k-p and y=0, we get

`k-p=0` ...(iii)

Adding (ii) and (iii), we get

`2k=2`

`k=1`

Putting k=1 in (ii).

`1+p=2`

`p=2-1`

`p=1`

Putting h=0, k=1 and p=1 in (i).

`y=(1)/(4(1))(x-(0))^2+(1)`

`y=(1)/(4)x^2+1`

Therefore, the equation of required parabola is
`y=(1)/(4)x^2+(1)`.