Answer
(a)
Vertex: $V(-\frac{1}{2},0)$
Focus: $F(0,p)=F(-\frac{1}{2},-\frac{1}{16})$
Directrix: $y=-p=\frac{1}{16}$
(b)
Work Step by Step
Equation of a parabola with vertical axis and vertex at $(h,k)$:
$(x-h)^2=4p(y-k)$
$-4(x+\frac{1}{2})^2=y$
$[x-(-\frac{1}{2})]^2=-\frac{1}{4}(y-0)$
$h=-\frac{1}{2}$
$k=0$
Vertex: $V(-\frac{1}{2},0)$
$4p=-\frac{1}{4}$
$p=-\frac{1}{16}$. Parabola opens downward.
The given equation can be obtained by shifting
$x^2=-\frac{1}{4}y$
left $\frac{1}{2}$ unit. In this equation:
Focus: $F(0,p)=F(0,-\frac{1}{16})$
Directrix: $y=-p=\frac{1}{16}$
Now, shift the vertex $\frac{1}{2}$ units to the left:
Focus: $F(0,p)=F(-\frac{1}{2},-\frac{1}{16})$
Directrix: $y=-p=\frac{1}{16}$
