SOLUTION: Find the focus, directrix, and equation of the parabola with vertex at the origin, axis along y-axis; opening upward; and the length of the latus rectum is 12. Sketch the graph.

Question 1170336: Find the focus, directrix, and equation of the parabola with vertex at the origin, axis along y-axis; opening upward; and the length of the latus rectum is 12. Sketch the graph.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's break down this problem step-by-step.
**1. Equation of the Parabola:**
* Since the vertex is at the origin (0, 0), the axis is along the y-axis, and the parabola opens upward, the equation of the parabola is of the form:
$x^2 = 4py$
where $p > 0$.
**2. Length of the Latus Rectum:**
* The length of the latus rectum is given as 12.
* The length of the latus rectum is also equal to $4p$.
* Therefore, $4p = 12$.
* Solving for $p$:
$p = \frac{12}{4} = 3$
**3. Equation of the Parabola (Specific):**
* Substitute $p = 3$ into the equation $x^2 = 4py$:
$x^2 = 4(3)y$
$x^2 = 12y$
**4. Focus:**
* The focus of the parabola is at (0, p).
* Since $p = 3$, the focus is at (0, 3).
**5. Directrix:**
* The directrix is a horizontal line given by the equation $y = -p$.
* Since $p = 3$, the directrix is $y = -3$.
**6. Sketch the Graph:**
* **Vertex:** (0, 0)
* **Focus:** (0, 3)
* **Directrix:** y = -3
* **Latus Rectum:** The latus rectum passes through the focus and is perpendicular to the axis of symmetry. Its endpoints are at a distance of $2p$ from the focus.
* Since $2p = 6$, the endpoints of the latus rectum are at (-6, 3) and (6, 3).
**Graph:**
```
^ y-axis
|
|
3 | * (0, 3) - Focus
| / \
| / \
|/ \
0 +-----------+---> x-axis
|\ /|
| \ / |
| \ / |
-3 | v - Directrix
|
```
**Summary:**
* **Equation of the Parabola:** $x^2 = 12y$
* **Focus:** (0, 3)
* **Directrix:** $y = -3$

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