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