Question

What is the vertex form of the equation of the parabola with a focus at (0,-15) and a directrix of `y=-16`?

Answer 1

The vertex form of a parabola is `y=a(x-h)+k`, but with what is given it is easier to start by looking at the standard form, `(x-h)^2=4c(y-k)`.

The vertex of the parabola is `(h,k)`, the directrix is defined by the equation `y=k-c`, and the focus is `(h,k+c)`. `a=1/(4c)`.

For this parabola, the focus `(h,k+c)` is `(0,"-"15)` so `h=0` and `k+c="-"15`.

The directrix `y=k-c` is `y="-"16` so `k-c="-"16`.

We now have two equations and can find the values of `k` and `c`:
`{(k+c="-"15),(k-c="-"16):}`

Solving this system gives `k=("-"31)/2` and `c=1/2`. Since `a=1/(4c)`, `a=1/(4(1/2))=1/2`

Plugging the values of `a`, `h`, and `k` into the first equation, we know the vertex form of the parabola is `y=1/2(x-0)+("-"31)/2`, or `y=1/2x-("-"31)/2`