Answer:
The vertex is (0,0), focus of the parabola is [tex](\frac{1}{16},0)[/tex], directrix of the parabola is [tex]y=-\frac{1}{16}[/tex], focal width is [tex]\frac{1}{4}[/tex].
Step-by-step explanation:
The given equation of parabola is
[tex]x=4y^2[/tex]
It can be written as
[tex]y^2=\frac{1}{4}x[/tex] ....(1)
The general equation of parabola is
[tex](y-k)^2=4p(x-h)[/tex] ... (2)
Where, (h,k) is vertex, (h+p,k) is focus, y=h-p is directrix and |4p| is focal width.
On comparing (1) and (2), we get
[tex]h=0,k=0[/tex]
The vertex is (0,0).
[tex]4p=\frac{1}{4}[/tex]
[tex]p=\frac{1}{16}[/tex]
Focus of the parabola is
[tex](h+p,k)=(0+\frac{1}{16},0)=(\frac{1}{16},0)[/tex]
Therefore focus of the parabola is [tex](\frac{1}{16},0)[/tex].
Directrix of the parabola is
[tex]y=h-p=0-\frac{1}{16}=-\frac{1}{16}[/tex]
Directrix of the parabola is [tex]y=-\frac{1}{16}[/tex].
Focal width is
[tex]|4p|=|4\times \frac{1}{16}|=\frac{1}{4}[/tex]
Focal width is [tex]\frac{1}{4}[/tex].
