Question

Demetrius is drafting a parabola on a coordinate plane. He plans to have the focus at (3,1) and the directrix at y=−3 . What is the equation in standard form that represents the parabola?

Answer 1

To find the equation of the parabola given a focus at \((3, 1)\) and a directrix at \(y = -3\), we can follow these steps:

1. **Identify the vertex of the parabola**:
The vertex lies midway between the focus and the directrix. Since the focus is at \( (3, 1) \) and the directrix is the line \(y = -3\), we can find the \(y\)-coordinate of the vertex by taking the average of the \(y\)-coordinate of the focus and the \(y\)-coordinate of the directrix:

\[
\text{Vertex } y = \frac{1 + (-3)}{2} = \frac{1 - 3}{2} = \frac{-2}{2} = -1
\]

The \(x\)-coordinate of the vertex is the same as the \(x\)-coordinate of the focus, so the vertex is at

\[
(3, -1).
\]

2. **Determine the distance from the vertex to the focus**:
The distance \(p\) from the vertex to the focus is \(1 - (-1) = 2\). Since the focus is above the vertex, the parabola opens upwards, and \(p = 2\).

3. **Using the standard form of the parabola's equation**:
The standard form for a parabola that opens upwards with vertex \((h, k)\) is given by:

\[
(x - h)^2 = 4p(y - k).
\]

Substituting \(h = 3\), \(k = -1\), and \(p = 2\):

\[
(x - 3)^2 = 4 \cdot 2 \cdot (y + 1).
\]

This simplifies to:

\[
(x - 3)^2 = 8(y + 1).
\]

4. **Final equation**:
We can rearrange the equation:

\[
(x - 3)^2 = 8y + 8.
\]

Thus, the equation of the parabola in standard form is:

\[
\boxed{(x - 3)^2 = 8(y + 1)}.
\]