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?

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:

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