How to find focus and directrix of $4y = x^2 -16$

Question

How would I go about finding the focus and directrix of $4y = x^2 -16$ $(1)$

I did the $SP^2 = QP^2$ and got $(y+b)^2 = (x-a)^2 + (y-b)^2$.

This gave me: $4by = x^2 -2ax +a^2$. Comparing this to $(1)$. I get $a = 0$ but I don't know what to do to get $b$. Can someone advise?

Edit: my workings:

Answer 1

Use definition of parabola as stated in comment:

$\sqrt{[(y-b)-a]^2+x^2}=(y-b)+a$

$(y-b)^2-2a(y-b)+a^2+x^2=(y-b)^2+2a(y-b)+a^2$

Which gives:

$4a(y-b)=x^2$

Comparing with given equation $x^2=4y+16$ we get:

$b=-4$ and $a=1$, where b is value of translation in y direction. The coordinates of vertex is (0, -4), therefore:

Equation of directrix is L: $y=-1-4=-5$.

Coordinate of focus is: $F(0, -3)$.

Answer 2

Let $(P): 4y=x^2-16$ be the parabola of focus $F$ and directrix $(d)$. By definition, the distance from any point $M$ on $(P)$ to the directrix $(d)$ equals the distance from $M$ to the focus $F$. Moreover, since $(P)$ is vertical and directed upwards, the focus is above the vertex and the directrix is horizontal and below it. We will use this in two steps.

1- Note that $(P)$ has a vertex $V(0,-4)$. This means that $F$ has a zero abcissa, say $(d):y=a$ and $F(0,b)$. Since $(d)$ is below the vertex and $F$ is above it, $a<-4$ and $b>-4$. As the vertex $V$ is a point on $(P)$, then:

$VF=distance(V$$\rightarrow$$(d))$. Meaning $|-4-b|=|-4-a|$. Which gives $b+4=-4-a$. This implies that $a=-b-8$.

This enables us to write $F(0,b)$ and $(d):y=-b-8$ where $b$ is to be determined.

2- Choose another point on $(P)$, say $M(4,0)$. Then:

$MF^2=distance(M$$\rightarrow$$(d))^2$. Meaning $(4-0)^2+(0-b)^2=(-b-8)^2$, which gives $b=-3$. This gives $a=-5$.

Hence the focus is $F(0,-3)$ and the directrix is$(d):y=-5$.