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Chapter 11: Problem 81

Lamp Shadow The light from a lamp creates a shadow on a wall with a hyperbolic border. Find the equation of the border if the distance between the vertices is 18 inches and the foci are 4 inches from the vertices. Assume the center of the hyperbola is at the origin.

Short Answer

Expert verified
\[ \frac{x^2}{81} - \frac{y^2}{88} = 1 \]

Step by step solution

01

Identify Hyperbola Properties

Given the vertices' distance of 18 inches, the distance from the center to each vertex (a) is half of this: \[ 2a = 18 \ a = 9 \]
02

Determine Distance to Foci

The foci are 4 inches away from the vertices, giving the distance from the center to each focus (c) as: \[ c = a + 4 = 9 + 4 = 13 \]
03

Calculate b

Using the relationship between the hyperbola's components: \[ c^2 = a^2 + b^2 \] Substitute the known values: \[ 13^2 = 9^2 + b^2 \ 169 = 81 + b^2 \ b^2 = 88 \ b = \sqrt{88} \]
04

Write Hyperbola Equation

The standard form of the hyperbola centered at the origin is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] Substituting the values of a and b gives: \[ \frac{x^2}{81} - \frac{y^2}{88} = 1 \]

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Key Concepts

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

hyperbola properties
A hyperbola is a type of conic section, formed by the intersection of a double cone with a plane that cuts through both halves of the cone. Each hyperbola consists of two separate curves called branches. These branches are mirror images of each other.

A few key properties of hyperbolas are:
  • They have two focal points (foci) that are outside of the curve
  • The transverse axis, which passes through the vertices and foci
  • The conjugate axis, perpendicular to the transverse axis

For the given exercise, the hyperbola's characteristics include the distances between the vertices (18 inches) and the foci (4 inches from the vertices).

Understanding these properties helps us derive the equation of the hyperbola and calculate its features.
distance to foci
The foci of a hyperbola are crucial points located along the transverse axis, a certain distance away from the center. The distance from the center to each focus is denoted as 'c'.

In our exercise, it is stated that the foci are 4 inches away from the vertices. Given that the vertices are 9 inches from the center (since 2a=18 inches and a=focal distance), we determine the distance 'c' using:

c = a + 4
c = 9 + 4
c = 13

By understanding the distance to the foci, we can further deduce other important parameters of the hyperbola.
vertices distance
The vertices of a hyperbola are the points where the two branches are closest to each other. The distance between the vertices is significant, as it helps to determine the hyperbola's parameters. For horizontal hyperbolas, the distance between the vertices is '2a'.

From our exercise, we know:

2a = 18 inches
a = 9 inches
This distance signifies how wide the hyperbola opens.

Knowing the vertices' distance and the distance to the foci, we can calculate 'b' using the formula:

c^{2} = a^{2} + b^{2}, where we already have a and b.

'b' represents the distance from the center to the asymptotes' intersection along the perpendicular axis.
hyperbola standard form
The standard form of the hyperbola equation is a clear representation of its shape and orientation. The general equation of a hyperbola centered at the origin is:

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

In this formula, 'a' represents the distance from the center to each vertex along the x-axis, and 'b' represents the distance relative to each corresponding vertex distance along the y-axis.

For our exercise, substituting the values of 'a' and 'b' we derived:

\[ a = 9 \rightarrow a^{2} = 81 \]
\[ b = \frac{\begin{90pt}\blackC{88} \rightarrow b^{2} = 88 \rightarrow \frac{x^2}{81} - \frac{y^2}{88} = 1\]

Thus, the equation for the hyperbolic border described in the exercise is:

\[ \frac{x^2}{81} - \frac{y^2}{88} = 1 \]
Always make sure to identify the proper parameters that feed into these forms to properly model hyperbolic curves.

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Most popular questions from this chapter

A satellite dish is shaped like a paraboloid of revolution. The signals that emanate from a satellite strike the surface of the dish and are reflected to a single point, where the receiver is located. If the dish is 10 feet across at its opening and 4 feet deep at its center, at what position should the receiver be placed?

A racetrack is in the shape of an ellipse 100 feet long and 50 feet wide. What is the width 10 feet from a vertex?

Find the midpoint of the line segment connecting the points (3,-8) and (-2,5)

Arch in St. Louis is often mistaken to be parabolic in shape. In fact, it is a catenary, which has a more complicated formula than a parabola. The Arch is 630 feet high and 630 feet wide at its base. (a) Find the equation of a parabola with the same dimensions. Let \(x\) equal the horizontal distance from the center of the arch. (b) The table below gives the height of the Arch at various widths; find the corresponding heights for the parabola found in (a). $$ \begin{array}{|cc|} \hline \text { Width (ft) } & \text { Height (ft) } \\ \hline 567 & 100 \\ 478 & 312.5 \\ 308 & 525 \\ \hline \end{array} $$ (c) Do the data support the notion that the Arch is in the shape of a parabola?

Hyperbolic Mirrors Hyperbolas have interesting reflective properties that make them useful for lenses and mirrors. For example, if a ray of light strikes a convex hyperbolic mirror on a line that would (theoretically) pass through its rear focus, it is reflected through the front focus. This property, and that of the parabola, were used to develop the Cassegrain telescope in \(1672 .\) The focus of the parabolic mirror and the rear focus of the hyperbolic mirror are the same point. The rays are collected by the parabolic mirror, then reflected toward the (common) focus, and thus are reflected by the hyperbolic mirror through the opening to its front focus, where the eyepiece is located. If the equation of the hyperbola is \(\frac{y^{2}}{9}-\frac{x^{2}}{16}=1\) and the focal length (distance from the vertex to the focus) of the parabola is \(6,\) find the equation of the parabola.
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