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Chapter 11: Problem 9

Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. \(x^{2}-y^{2}=1\)

Short Answer

Expert verified
The vertices are \((\pm 1, 0)\); foci are \((\pm \sqrt{2}, 0)\); asymptotes are \(y = \pm x\).

Step by step solution

01

Identify the Standard Form of a Hyperbola

The given equation is \(x^2 - y^2 = 1\). Compare this with the standard form of a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Here, \(a^2 = 1\) and \(b^2 = 1\), hence, \(a = 1\) and \(b = 1\).
02

Determine the Center

For the equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the center of the hyperbola \((h, k)\) is determined from the lack of terms \(x-h\) and \(y-k\). Here, the center is at the origin \((0, 0)\).
03

Find the Vertices

The vertices of a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are at \((h \pm a, k)\). With \(a = 1\) and center \((0, 0)\), the vertices are \((\pm 1, 0)\).
04

Find the Foci

The foci of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are located at \((h \pm c, k)\) where \(c = \sqrt{a^2 + b^2}\). Calculating \(c = \sqrt{1 + 1} = \sqrt{2}\), the foci are at \((\pm \sqrt{2}, 0)\).
05

Determine the Asymptotes

For the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the equations of the asymptotes are \(y = \pm \frac{b}{a}x\). With \(a = 1\) and \(b = 1\), the asymptotes are \(y = \pm x\).
06

Sketch the Hyperbola

Sketching involves plotting the center \((0, 0)\), the vertices \((\pm 1, 0)\), and the foci \((\pm \sqrt{2}, 0)\). The asymptotes, \(y = x\) and \(y = -x\), should also be drawn as diagonal lines through the origin forming a 'X'. Sketch the hyperbola approaching these asymptotes.

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Key Concepts

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

Vertices of Hyperbola
The vertices of a hyperbola are crucial points. They provide the tips of the hyperbola's "open ends." In the standard form, \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the vertices are positioned at \((h \pm a, k)\). Here, \(h\) and \(k\) represent the center of the hyperbola.

Given our equation \(x^2 - y^2 = 1\), which can be rewritten as \(\frac{x^2}{1} - \frac{y^2}{1} = 1\), we find that both \(a^2\) and \(b^2\) are 1. Thus, \(a = 1\) and the center is at origin \( (0, 0) \).
  • This means the vertices are at \((0 \pm 1, 0)\), simplifying to \((1, 0)\) and \((-1, 0)\).
These vertices help define the shape and size of the hyperbola. In our case, they lie along the x-axis.
Foci of Hyperbola
The foci of a hyperbola play a similar role to the vertices, marking points inside each "branch" of the hyperbola, pulling the curve outward. For identifying the foci, you need a number \(c\) determined by the formula \(c = \sqrt{a^2 + b^2}\).

For the equation \(x^2 - y^2 = 1\), we have \(a^2 = 1\) and \(b^2 = 1\). By plugging into the formula, \(c = \sqrt{1 + 1} = \sqrt{2}\). The foci, based on the formula, are situated at \((h \pm c, k)\).
  • Thus, the foci positions are \((0 \pm \sqrt{2}, 0)\), simplifying to \((\sqrt{2}, 0)\) and \((-\sqrt{2}, 0)\).
These points might not be visible in simple graphs, but they are vital when understanding the hyperbola's properties and its stretched shape along the x-axis.
Asymptotes of Hyperbola
Asymptotes are lines that the hyperbola approaches but never actually touches. They provide a "frame" for the hyperbola, indicating the direction in which the branches of the hyperbola extend.

For a hyperbola in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the equations of the asymptotes are \(y = \pm \frac{b}{a}x\).
  • Using our values, \(a = 1\) and \(b = 1\), the equations simplify to \(y = \pm x\).
These lines pass through the center of the hyperbola at the origin \( (0, 0) \) and extend diagonally. When sketching, ensure that these asymptotes form a cross-like "X" to guide the shape of the hyperbola branches.

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Most popular questions from this chapter

Find the vertices, foci, and eccentricity of the ellipse. Determine the lengths of the major and minor axes, and sketch the graph. $$ 4 x^{2}+25 y^{2}=100 $$

(a) Use the discriminant to identify the conic. (b) Confirm your answer by graphing the conic using a graphing device. $$2 x^{2}-4 x y+2 y^{2}-5 x-5=0$$

Find an equation for the ellipse that satisfies the given conditions. Length of major axis \(10,\) foci on \(x\) -axis, ellipse passes through the point \((\sqrt{5}, 2)\)

\begin{array}{l}{1-22 \text { a pair of parametric equations is given. }} \\\ {\text { (a) Sketch the curve represented by the parametric equations. }} \\\ {\text { (b) Find a rectangular-coordinate equation for the curve by }} \\\ {\text { eliminating the parameter. }}\end{array} $$ x=t^{2}, \quad y=t^{4}+1 $$

(a) Use a graphing device to sketch the top half (the portion in the first and second quadrants) of the family of ellipses \(x^{2}+k y^{2}=100\) for \(k=4,10,25,\) and \(50 .\) (b) What do the members of this family of ellipses have in common? How do they differ?
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