Test IV Problem Solving.
A certain bridge is supported by a reinforced parabolic arc. The bridge deck , being the
directrix of the parabolic are, has the equation y=24 where 24 represents its vertical
clearance of 24 ft from the surface of the river.The bridge deck is 6 ft above the peak of
the are. If the tallest column on the left represents the y-axis and is 60 ft away from the
corresponding column to the right find the coordinates of the vertex, the focus,the endpoints
of the latus rectum, and the general equation that describes the arc.

To solve this problem, let's break it down step by step.

### Problem Restatement:
We have a parabolic arc supporting a bridge. The directrix of the parabola is the bridge deck, given by the equation \( y = 24 \). The parabola's vertex is 6 feet below the deck, and the total horizontal span of the bridge (distance between the leftmost and rightmost columns) is 60 feet.

### Step 1: Determine the Coordinates of the Vertex
Given that the deck (directrix) is at \( y = 24 \) and the vertex is 6 feet below this, the vertex \( V \) of the parabola will be at:
\[
y_V = 24 - 6 = 18
\]
The parabola is symmetric about the y-axis, so the x-coordinate of the vertex is \( 0 \). Therefore, the coordinates of the vertex are:
\[
V = (0, 18)
\]

### Step 2: Determine the Focus
The distance from the vertex to the directrix is equal to the distance from the vertex to the focus \( F \). Let the distance between the vertex and the focus be \( p \).

The directrix is at \( y = 24 \), and the vertex is at \( y = 18 \). Therefore:
\[
p = 24 - 18 = 6 \text{ feet}
\]

Since the parabola opens downwards (vertex is above the focus), the focus is located \( p = 6 \) feet below the vertex:
\[
y_F = 18 - 6 = 12
\]
So, the coordinates of the focus are:
\[
F = (0, 12)
\]

### Step 3: Find the General Equation of the Parabola
The standard equation for a parabola that opens vertically (upwards or downwards) is:
\[
(y - k) = \frac{1}{4p}(x - h)^2
\]
Where \( (h, k) \) is the vertex, and \( p \) is the distance from the vertex to the focus.

Substituting \( (h, k) = (0, 18) \) and \( p = 6 \):
\[
(y - 18) = \frac{1}{24}x^2
\]
Or:
\[
y = \frac{1}{24}x^2 + 18
\]

### Step 4: Determine the Endpoints of the Latus Rectum
The latus rectum is a line segment through the focus that is parallel to the directrix and passes through the parabola. The length of the latus rectum is \( 4p \).

Here, \( p = 6 \), so the length of the latus rectum is:
\[
4p = 24 \text{ feet}
\]

The endpoints of the latus rectum lie at equal distances horizontally from the focus. Since the length is 24 feet, each endpoint is 12 feet to the left and right of the focus.

So, the endpoints of the latus rectum are:
\[
\left(-12, 12\right) \quad \text{and} \quad \left(12, 12\right)
\]

### Final Summary:
- **Vertex:** \( V = (0, 18) \)
- **Focus:** \( F = (0, 12) \)
- **Endpoints of the Latus Rectum:** \( (-12, 12) \) and \( (12, 12) \)
- **General Equation of the Parabola:** \( y = \frac{1}{24}x^2 + 18 \)

Would you like any more details or have further questions?

---

Here are some related questions you might find interesting:
1. How would the equation of the parabola change if the vertex was moved 2 feet higher?
2. What would be the effect on the focus if the bridge deck was 30 feet above the river?
3. How do you find the vertex and focus if given the general form \( ax^2 + bx + c = y \)?
4. What is the significance of the latus rect

Explore how to solve problems involving parabolic bridges. Learn the coordinates of the vertex and focus, endpoints of the latus rectum, and derive the general equation of the parabola with detailed step-by-step solutions.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation