SOLUTION: Find the length of the latus rectum and the equation of the parabola with vertex at the origin, directrix x=-3 and focus (3, 0). Sketch the graph.

Question 1170335: Find the length of the latus rectum and the equation of the parabola with vertex at the origin, directrix x=-3 and focus (3, 0). Sketch the graph.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's solve this problem step-by-step.
**1. Determine the Orientation of the Parabola:**
* The vertex is at the origin (0, 0).
* The focus is at (3, 0).
* The directrix is x = -3.
* Since the focus is to the right of the vertex and the directrix is to the left, the parabola opens to the right.
**2. Determine the Value of 'p':**
* The distance between the vertex and the focus is 'p'.
* The distance between the vertex (0, 0) and the focus (3, 0) is 3.
* Therefore, p = 3.
**3. Find the Equation of the Parabola:**
* Since the parabola opens to the right and the vertex is at the origin, the equation is of the form:
$y^2 = 4px$
* Substitute p = 3 into the equation:
$y^2 = 4(3)x$
$y^2 = 12x$
**4. Find the Length of the Latus Rectum:**
* The length of the latus rectum is 4p.
* Since p = 3, the length of the latus rectum is 4(3) = 12.
**5. Sketch the Graph:**
* **Vertex:** (0, 0)
* **Focus:** (3, 0)
* **Directrix:** x = -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 (3, 6) and (3, -6).
**Graph:**
```
^ y-axis
|
6 | * (3, 6)
| /
| /
| /
| /
0 +-----------+---> x-axis
| \
| \
| \
-6 | * (3, -6)
|
-3 | Directrix x=-3
```
**Summary:**
* **Length of the Latus Rectum:** 12
* **Equation of the Parabola:** $y^2 = 12x$

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