Given:
Focus of a parabola = (0,2)
Directrix: y=0.
To find:
The equation of parabola.
Solution:
The equation of parabola is
[tex]y=\dfrac{1}{4p}(x-h)^2+k[/tex] ...(i)
where, (h,k) is vertex, (h,k+p) is focus, y=k-p is directrix.
Focus : [tex](h,k+p)=(0,2)[/tex]
On comparing both sides, we get
[tex]h=0[/tex]
[tex]k+p=2[/tex] ...(ii)
On comparing y=k-p and y=0, we get
[tex]k-p=0[/tex] ...(iii)
Adding (ii) and (iii), we get
[tex]2k=2[/tex]
[tex]k=1[/tex]
Putting k=1 in (ii).
[tex]1+p=2[/tex]
[tex]p=2-1[/tex]
[tex]p=1[/tex]
Putting h=0, k=1 and p=1 in (i).
[tex]y=\dfrac{1}{4(1)}(x-(0))^2+(1)[/tex]
[tex]y=\dfrac{1}{4}x^2+1[/tex]
Therefore, the equation of required parabola is [tex]y=\dfrac{1}{4}x^2+(1)[/tex].
