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Chapter 10: Problem 36

Spotlight. \(\quad\) A spotlight has a parabolic cross section that is \(4 \mathrm{ft}\) wide at the opening and \(1.5 \mathrm{ft}\) deep at the vertex. How far from the vertex is the focus?

Short Answer

Expert verified
The focus is approximately 0.67 ft from the vertex.

Step by step solution

01

- Write the General Equation of a Parabola

The general equation of a parabola with a vertical axis is given by \( y = ax^2 \). The vertex form of the equation is useful as well, \( y = a(x-h)^2 + k \), where \( (h,k) \) is the vertex. For this problem, we can place the vertex at the origin (0,0) for simplicity, so \( y = ax^2 \).
02

- Determine Parameters from Given Dimensions

The width at the opening is 4 ft, making the span from -2 ft to 2 ft horizontally (since width \ 4ft \ means \ 2ft \ either side of the y-axis). The depth of 1.5 ft means when \( x = 卤2 \), \( y = 1.5 \). Substitute these points into the equation \ y = ax^2 \ to find 'a'.
03

- Solve for Coefficient 'a'

Substitute \( x=2 \) and \( y=1.5 \): y: 1.5 = a(2^2) \Rightarrow 1.5 = 4a \Rightarrow a = \frac{1.5}{4} = 0.375 Therefore, the equation of the parabola is \( y = 0.375x^2 \).
04

- Find the Focus of the Parabola

The focus of a parabola \( y = ax^2 \) is at distance \( \frac{1}{4a} \) from the vertex along the axis of symmetry. Here, \ a = 0.375 \.\frac{1}{4a} = \frac{1}{4(0.375)} = \frac{1}{1.5} = \frac{2}{3} \ ft. Thus, the focus is 0.67 ft from the vertex.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

parabola equation
A parabola is a distinct U-shaped curve that can be defined mathematically. The standard form of a parabolic equation with a vertical orientation is given by \( y = ax^2 \).

In this equation, `a` is a coefficient that determines how
vertex form of a parabola
One of the useful representations of a parabola is its vertex form, given by \( y = a(x-h)^2 + k \), where \((h,k)\) is the vertex of the parabola.
The vertex is the turning point of the parabola and gives a clear indication of its maximum or minimum point.

For our spotlight problem, we can simplify things by assuming the vertex is at the origin \((0,0)\), making the equation look like \( y = ax^2 \). This setup eases the calculation, as it simplifies the determination of the coefficient and the focus.
focus of a parabola
A fascinating property of a parabola is its focus. The focus of a parabola is a specific point where light rays or signals will converge after reflecting off the parabolic surface.

For the equation \( y = ax^2 \), the distance from the vertex to the focus is given by \(\frac{1}{4a}\).
This formula is derived from the geometric properties of the parabola and is critical in problems involving light or sound reflection.
In our spotlight problem, with \( a = 0.375 \), the distance to the focus is calculated as:
\ \frac{1}{4(0.375)} = \frac{1}{1.5} = \frac{2}{3} \ft \.
Thus the focus is 0.67 ft from the vertex.
coefficient determination
Finding the coefficient `a` is a crucial step to fully defining the parabola's shape. This can be done by substituting known values into the equation \( y = ax^2 \).
In the spotlight problem, the opening is 4 ft wide, leading to horizontal values of \( x = 卤2 \) ft and a depth of 1.5 ft.
By substituting these values, \( y = 1.5 \) at \( x = 2 \):
\ 1.5 = a(2^2) \
Which simplifies to:
\ a = \frac{1.5}{4} = 0.375 \.
This calculated coefficient is then used to identify the complete parabolic equation \( y = 0.375x^2 \).

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