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Chapter 7: Problem 56

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

Short Answer

Expert verified
The graph of the given equation is a hyperbola centered at (0,0) opening along the x-axis with vertices at (卤3,0) and co-vertices at (0,卤1). Its asymptotes are the lines \(y=卤\frac{1}{3}x\).

Step by step solution

01

Identifying Components

Identify the center, major axis and vertices from the given equation. In this case, the center is at the origin (0,0) because there's no translation in the equation. The a value or the semi-major axis is \(\sqrt{9}=3\) in the x direction and the b value or the semi-minor axis is \(\sqrt{1}=1\) in the y direction. Hence, the vertices are at points (卤3,0).
02

Identify Co-Verticies

The co-vertices of the hyperbola can also be determined from the given equation by moving a distance 'b' in the y direction from the center. Hence, the co-vertices are at points (0,卤1).
03

Draw Asymptotes

The asymptotes of the hyperbola are the diagonals of the rectangle formed by vertices and co-vertices. The equation of asymptotes in this case will be \(y=卤\frac{b}{a}x=卤\frac{1}{3}x\). Plot these lines on graph.
04

Sketch the Hyperbola

Since this is a horizontal hyperbola (as the x term is positive), draw the hyperbola opening left and right through the vertices, approaching the asymptotes as it moves away from the center.

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Most popular questions from this chapter

Graph each ellipse and give the location of its foci. $$\frac{(x-4)^{2}}{9}+\frac{(y+2)^{2}}{25}=1$$

Graph each ellipse and give the location of its foci. $$\frac{(x-2)^{2}}{9}+\frac{(y-1)^{2}}{4}=1$$

In Exercises \(1-18,\) graph each ellipse and locate the foci. $$25 x^{2}+4 y^{2}=100$$

Find the standard form of the equation of each ellipse satisfying the given conditions. $$\text { Foci: }(0,-3),(0,3) ; \text { vertices: }(0,-4),(0,4)$$

In Exercises \(1-18,\) graph each ellipse and locate the foci. $$\frac{x^{2}}{9}+\frac{y^{2}}{36}=1$$
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