Answer
(a)
Vertex: $V(\frac{1}{2},0)$
Focus: $F(\frac{9}{2},0)$
Directrix: $x=-\frac{7}{2}$
(b)
Work Step by Step
Equation of a parabola with horizontal axis and vertex at $(h,k)$:
$(y-k)^2=4p(x-h)$
$y^2=16x-8$
$y^2=16(x-\frac{1}{2})$
$(y-0)^2=16(x-\frac{1}{2})$
$h=\frac{1}{2}$
$k=0$
Vertex: $V(\frac{1}{2},0)$
$4p=16$
$p=4$. Parabola opens to the right.
The given equation can be obtained by shifting
$y^2=16x$
right $\frac{1}{2}$ unit:
Focus: $F(p,0)=F(4,0)$
Directrix: $x=-p=-4$
Now, shift the vertex and the directrix $\frac{1}{2}$ unit to the right:
Focus: $F(\frac{9}{2},0)$
Directrix: $x=-\frac{7}{2}$
