Question

Olivia is asked to graph a parabola on a coordinate plane.The focus is located at $(0,-3)$ , and the directrix is at
$y=1$ . Fill in the missing values on the table for the coordinates of the parabola (1 point)
\begin{tabular}{|c|c|}
\hline
\( x \)-value & \( y \)-value \\
\hline
2 & -4 \\
\hline
0 & \\
\hline
-1 & -\frac{7}{4} \\
\hline
& -\frac{4}{3} \\
\hline
\end{tabular}

Answer 1

To find the equation of the parabola, we start with the focus \((0, -3)\) and directrix \(y = 1\). The parabola opens downwards because the focus is below the directrix.

The standard equation of a parabola with a vertical axis of symmetry is given by:
\[
(x - h)^2 = 4p(y - k)
\]
where the vertex \((h, k)\) is halfway between the focus and directrix, and \(p\) is the distance from the vertex to the focus.

1. **Finding the vertex**:
The focus is at \((0, -3)\), and the directrix is at \(y = 1\). The vertex is the midpoint between the focus and directrix:
\[
k = \frac{-3 + 1}{2} = \frac{-2}{2} = -1 \quad \text{and} \quad h = 0
\]
Therefore, the vertex is \((0, -1)\).

2. **Finding \(p\)**:
The distance from the vertex to the focus (or directrix) is:
\[
p = -1 - (-3) = 2
\]
Since the parabola opens downwards, we take \(p = -2\) (negative value).

3. **Equation of the parabola**:
Plugging \(h = 0\), \(k = -1\), and \(p = -2\) into the equation:
\[
(x - 0)^2 = 4(-2)(y + 1)
\]
Simplifying, we get:
\[
x^2 = -8(y + 1)
\]
Rearranging gives:
\[
y = -\frac{1}{8}x^2 - 1
\]

Now we can calculate the missing \(y\)-values for the given \(x\)-values using this equation.

1. When \(x = 0\):
\[
y = -\frac{1}{8}(0)^2 - 1 = -1
\]

2. For \(x = -1\):
\[
y = -\frac{1}{8}(-1)^2 - 1 = -\frac{1}{8} - 1 = -\frac{1}{8} - \frac{8}{8} = -\frac{9}{8}
\]

3. To find the missing \(x\)-value for \(y = -\frac{4}{3}\):
We set up the equation:
\[
-\frac{4}{3} = -\frac{1}{8}x^2 - 1
\]
Rearranging gives:
\[
-\frac{4}{3} + 1 = -\frac{1}{8}x^2
\]
Simplifying gives:
\[
-\frac{4}{3} + \frac{3}{3} = -\frac{1}{8}x^2 \implies -\frac{1}{3} = -\frac{1}{8}x^2
\]
Eliminating the negative sign gives:
\[
\frac{1}{3} = \frac{1}{8}x^2
\]
Multiplying both sides by 24:
\[
8 = 3x^2 \implies x^2 = \frac{8}{3} \implies x = \pm \sqrt{\frac{8}{3}} = \pm \frac{2\sqrt{2}}{\sqrt{3}} = \pm \frac{2\sqrt{6}}{3}
\]

4. Therefore, the table is filled as follows:

\[
\begin{tabular}{|c|c|}
\hline
\( x \)-value & \( y \)-value \\
\hline
2 & -4 \\
\hline
0 & -1 \\
\hline
-1 & -\frac{9}{8} \\
\hline
\frac{2\sqrt{6}}{3} & -\frac{4}{3} \\
\hline
\end{tabular}
\]

This gives the completed table with the missing \(y\)-values filled in.