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Chapter 7: Problem 68

Describe how to graph \(\frac{x^{2}}{9}-\frac{y^{2}}{1}=1\)

Short Answer

Expert verified
The graph of \(\frac{x^{2}}{9}-\frac{y^{2}}{1}=1\) is a hyperbola with center at (0, 0), vertices at (-3, 0) and (3, 0), co-vertices at (0, -1) and (0, 1), and asymptotes at y = 卤 (\( 1 / 3 \)) x. It opens to the right and left.

Step by step solution

01

Identify the Center

The center of the hyperbola is at the point (h, k) where h and k are each 0 in this case, because there is no addition or subtraction of a number with \(x^{2}\) or \(y^{2}\) in the equation. Thus, the center is at (0, 0).
02

Identify the Vertices

Vertices are found on the hyperbola, equidistant from the center. The distance to the vertices from the center in a horizontal hyperbola is given by the square root of the denominator of the x term. Here, the square root of 9 is 3. The vertices are then at (0卤3, 0) or (-3, 0) and (3, 0).
03

Identify the Co-vertices

Co-vertices are found on the perpendicular axis of symmetry. The y term gives the distance from the center to the co-vertices. Here, the square root of 1 is 1. Thus, the co-vertices are at (0, 卤1) or (0, -1) and (0, 1).
04

Draw the Asymptotes

Asymptotes of a hyperbola pass through its center and have slopes given by 卤 (\( \text{denominator y term} / \text{denominator x term} \)). So, the equations of the asymptotes are y = 卤 (\( 1 / 3 \)) x. Draw lines representing these asymptotes.
05

Draw the Hyperbola

Now, draw the two parts of the hyperbola on either side of the center, with edges that approach the asymptote lines as they move away from the center. The vertices and co-vertices should be on the hyperbola. Its orientation is horizontal, you open your hyperbola right and left.

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