Problem 16 - College Algebra

Question

Use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$\frac{x^{2}}{144}-\frac{y^{2}}{81}=1$$

Step-by-Step Solution

Answer

The hyperbola with the given equation has vertices at (-12, 0) and (12, 0), co-vertices at (0, -9) and (0, 9), foci at (-15, 0) and (15, 0) and asymptotes at \( y = ±\frac{3}{4}x \).

Step 1: Identify a, b and c

First, identify values of a, b in the given equation \( \frac{x^{2}}{144}-\frac{y^2}{81} = 1 \). Here a = \( \sqrt{144} = 12 \) and b = \( \sqrt{81} = 9 \). Then use these values to find c by using the relationship \( c = \sqrt{a^2 + b^2} = \sqrt{144 + 81} = 15 \) which represents distance from the center to a focus.

Step 2: Identify vertices, co-vertices and foci

The center of the hyperbola is located at (0,0) as there are no shifts from x and y in the equation. The vertices are a units left and right of the center. Therefore, the vertices are (-12, 0) and (12, 0). The co-vertices are b units up and down of the center, hence they are (0, -9) and (0, 9). The foci are (-15, 0) and (15, 0) as the value of c gives distance from center to a focus.

Step 3: Graph and draw asymptotes

Plot the vertices, co-vertices and foci. Draw the hyperbola passing through the vertices and approaching your asymptote lines as x approaches ±∞. The asymptotes for the hyperbola are the lines that intersect the center and pass through the corners of the rectangle formed by vertices and co-vertices. They act as guides showing how the hyperbola approaches infinity. The equations of the asymptotes for a horizontal hyperbola are \( y = k ± \frac{b}{a}(x - h) \). For this case, h = k = 0 yielding asymptotes as \( y = ± \frac{9}{12}x = ±\frac{3}{4}x \)

Key Concepts

Understanding Vertices of a Hyperbola

In a hyperbola, the vertices play a crucial role in defining the shape of its graph. The vertices are the points where the hyperbola intersects its transverse axis, which is the primary axis along which the hyperbola stretches. In the given equation \( \frac{x^2}{144} - \frac{y^2}{81} = 1 \), the transverse axis is horizontal.
Given that the equation has the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices can be found at a distance of \( a \) from the center along the x-axis. Here, \( a = 12 \).

• The center of the hyperbola is at \((0,0)\), making the vertices \((-12, 0)\) and \((12, 0)\).
• The distance between the vertices is \(2a = 24\).

The vertices, therefore, lay the foundation for sketching the basic framework of the hyperbola.

Asymptotes and Their Role

Asymptotes are imaginary lines that provide a framework guiding the approach of the hyperbola's arms as they extend towards infinity. For a hyperbola in the format \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the asymptotes are straight lines that meet at the hyperbola's center. They do not intersect the hyperbola but steer its trajectory.
The asymptotes of the given hyperbola are derived and have the equations:

• \( y = \pm \frac{3}{4}x \)

These slant through the origin, indicating how the hyperbola stretches and flares away to infinity. As a visual guide, they form an ‘X’ shape, effectively defining the limits for the hyperbola's arms.

Exploring the Foci

The foci of a hyperbola are pivotal points. They reside on the transverse axis, further from the center than the vertices, playing a major role in its reflective properties. For the hyperbola's equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the distance to each focus is \( c \), calculated using \( c = \sqrt{a^2 + b^2} \).

• In this case, \( c = 15 \).
• The foci are positioned symmetrically at \((-15, 0)\) and \((15, 0)\).

The hyperbola appears to "open" around these foci points, demonstrating a structural harmony that extends indefinitely towards them but never reaching them physically.

Finding the Equations of Asymptotes

The equations of the asymptotes determine their slope and direction, crucial for defining the hyperbola's arms' spread. In a hyperbola of form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), these lines take the form:
\[ y = k \pm \frac{b}{a}(x - h) \]
Here, \( h \) and \( k \) are the coordinates of the center, \((0,0)\) in this context.

• By substituting \( b = 9 \) and \( a = 12 \), the equations of the asymptotes become \( y = \pm \frac{3}{4}x \).

Understanding these equations helps in manually sketching the hyperbola, guiding its curve as it progresses towards infinity without reaching its asymptotes.