An arch in a memorial park, having a parabolic shape, has a base width of 28 feet and opens downward 
with the vertex at the origin. Suppose the base serves as its latus rectum. Find the equation (in standard 
form) that models this shape, then graph it. State the focus, directrix, and axis of symmetry. 

The graph shows the parabola \(x^2 = -28y\), which represents the shape of the arch. Here's a summary of the key features:

- **Focus**: \((0, -7)\)
- **Directrix**: \(y = 7\)
- **Axis of Symmetry**: \(x = 0\) (the y-axis)

These features align with the given conditions that the parabola opens downward with its vertex at the origin and has a latus rectum of 28 feet.

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### Related Questions:
1. How do you find the length of the latus rectum for any parabola?
2. What are the steps to derive the equation of a parabola given its vertex and focus?
3. How do you determine the direction in which a parabola opens?
4. Can you explain how the directrix and focus are related to the properties of a parabola?
5. What is the general form of a parabola's equation, and how does it differ from the vertex form?

**Tip:** Understanding the relationship between the latus rectum, focus, and directrix helps in solving various problems involving parabolas.

An arch in a memorial park, having a parabolic shape, has a base width of 28 feet and opens downward with the vertex at the origin. Suppose the base serves as its latus rectum. Find the equation (in standard form) that models this shape, then graph it. State the focus, directrix, and axis of symmetry.

Learn how to find the equation of a parabolic arch with a base width of 28 feet and its latus rectum as the base. This problem involves key concepts such as conic sections, focus, directrix, and axis of symmetry. Ideal for high school students in Grades 10-12.

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