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

Mar 27, 2017

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

The vertex of the parabola is $\left(h , k\right)$, the directrix is defined by the equation $y = k - c$, and the focus is $\left(h , k + c\right)$. $a = \frac{1}{4 c}$.

For this parabola, the focus $\left(h , k + c\right)$ is $\left(0 , \text{-} 15\right)$ so $h = 0$ and $k + c = \text{-} 15$.

The directrix $y = k - c$ is $y = \text{-} 16$ so $k - c = \text{-} 16$.

We now have two equations and can find the values of $k$ and $c$:
$\left\{\begin{matrix}k + c = \text{-"15 \\ k-c="-} 16\end{matrix}\right.$

Solving this system gives $k = \frac{\text{-} 31}{2}$ and $c = \frac{1}{2}$. Since $a = \frac{1}{4 c}$, $a = \frac{1}{4 \left(\frac{1}{2}\right)} = \frac{1}{2}$

Plugging the values of $a$, $h$, and $k$ into the first equation, we know the vertex form of the parabola is $y = \frac{1}{2} \left(x - 0\right) + \frac{\text{-} 31}{2}$, or $y = \frac{1}{2} x - \frac{\text{-} 31}{2}$