Given:
The given equation is
[tex]y-3=(2-x)^2[/tex]
To find:
The vertex, focus, and directrix.
Solution:
The equation of a parabola is
[tex]y-k=a(x-h)^2[/tex] ...(i)
where, (h,k) is vertex, [tex]\left(h,k+\dfrac{1}{4a}\right)[/tex] and directrix is [tex]y=k-\dfrac{1}{4a}[/tex]
We have,
[tex]y-3=(2-x)^2[/tex]
It can be written as
[tex]y-3=(-(x-2))^2[/tex]
[tex]y-3=(x-2)^2[/tex] ...(ii)
On comparing (i) and (ii), we get
[tex]h=2,k=3,a=1[/tex]
Vertex of the parabola is (2,3).
[tex]Focus=\left(2,3+\dfrac{1}{4(1)}\right)[/tex]
[tex]Focus=\left(2,3+\dfrac{1}{4}\right)[/tex]
[tex]Focus=\left(2,\dfrac{13}{4}\right)[/tex]
Therefore, the focus of the parabola is [tex]\left(2,\dfrac{13}{4}\right)[/tex].
Directrix of the parabola is
[tex]y=3-\dfrac{1}{4(1)}[/tex]
[tex]y=3-\dfrac{1}{4}[/tex]
[tex]y=\dfrac{11}{4}[/tex]
Therefore, the directrix of the parabola is [tex]y=\dfrac{11}{4}[/tex].
