Question

write the equation in vertex form for the parabola with focus (0, -9) and directrix y = -1. simplify any fractions.

Explanation:

Step1: Find the vertex

The vertex of a parabola is the mid - point between the focus \((0,-9)\) and the directrix \(y = - 1\).
The formula for the \(y\) - coordinate of the vertex \(k\) is \(k=\frac{-9+( - 1)}{2}=\frac{-9 - 1}{2}=\frac{-10}{2}=-5\). The \(x\) - coordinate of the vertex \(h = 0\) (since the focus and directrix are symmetric about the \(x = 0\) line). So the vertex is \((h,k)=(0,-5)\).

Step2: Find the value of \(p\)

The distance \(p\) between the vertex and the focus (or the vertex and the directrix).
Since the vertex is \((0,-5)\) and the focus is \((0,-9)\), then \(p=-9-( - 5)=-4\) (we can also calculate using the directrix: \(p=-5-( - 1)=-4\))

Step3: Write the vertex form of the parabola equation

The vertex form of a parabola is \(y=a(x - h)^{2}+k\), and for a parabola \(y=a(x - h)^{2}+k\), \(a=\frac{1}{4p}\)
Substitute \(h = 0\), \(k=-5\) and \(p=-4\) into the formula. First, \(a=\frac{1}{4\times(-4)}=-\frac{1}{16}\)
The equation of the parabola is \(y=-\frac{1}{16}(x - 0)^{2}-5\)

Answer:

\(y=-\frac{1}{16}x^{2}-5\)