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Chapter 6: Problem 19

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Then sketch the hyperbola using the asymptotes as an aid. $$x^{2}-y^{2}=1$$

Short Answer

Expert verified
The center of the hyperbola is at (0,0), vertices at (卤1,0), foci at (卤鈭�2, 0) and the equations of the asymptotes are \(y = \pm x\).

Step by step solution

01

Determine the Center

The center of the hyperbola is at the origin (0,0) because the equation is in the form \(x^{2} - y^{2} = 1\), which indicates that the hyperbola is centered at the origin.
02

Identify the Vertices

The vertices of the hyperbola are given by \((\pm a, 0)\) where \(a^2\) is the coefficient of \(x^2\). In this case, \(a^2 = 1\), so the vertices are at (卤1,0).
03

Compute the Foci

The foci of a hyperbola are at \((\pm c, 0)\) where \(c^2 = a^2 + b^2\). As there is no \(y^2\) term, \(b^2 = 1\). Therefore \(c^2 = 1 + 1 = 2\) , so the foci are at (卤鈭�2, 0).
04

Find the Asymptotes Equation

The equations of the asymptotes of the hyperbola are given by \(y = \pm (b/a)x\). As \(a^2 = 1\) and \(b^2 = 1\), the equations of the asymptotes are \(y = \pm x\).
05

Sketching the Hyperbola

Sketch the hyperbola on a graph using the center, vertices, foci, and asymptotes. The vertices serve as points on the inside edge of the hyperbola and the asymptotes give the outer boundary lines.

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Most popular questions from this chapter

In Exercises \(91-116\), convert the polar equation to rectangular form. $$r=\frac{2}{1+\sin \theta}$$

In Exercises \(91-116\), convert the polar equation to rectangular form. $$r=\frac{1}{1-\cos \theta}$$

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Explain how the graph of each conic differs from the graph of \(\left.r=\frac{5}{1+\sin \theta} . \text { (See Exercise } 17 .\right)\) (a) \(r=\frac{5}{1-\cos \theta}\) (b) \(r=\frac{5}{1-\sin \theta}\) (c) \(r=\frac{5}{1+\cos \theta}\) (d) \(r=\frac{5}{1-\sin [\theta-(\pi / 4)]}\)

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