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Chapter 11: Problem 35

Give information about the foci, vertices, and asymptotes of hyperbolas centered at the origin of the \(x y\) -plane. In each case, find the hyperbola's standard-form equation from the information given. Foci: \(\quad(0, \pm \sqrt{2})\) Asymptotes: \(\quad y=\pm x\)

Short Answer

Expert verified
The equation of the hyperbola is \(y^2 - x^2 = 1\).

Step by step solution

01

Recognizing Hyperbola Orientation

The provided foci \((0, \pm \sqrt{2})\) indicate that the hyperbola is vertical because these points are arranged along the \(y\)-axis. Additionally, the asymptotes \(y = \pm x\) also confirm that the hyperbola is centered at the origin and is vertical.
02

Standard Form of Hyperbola Equation

The standard form of the equation of a vertical hyperbola centered at the origin is given by:\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \] where \(a\) and \(b\) are constants related to the hyperbola's axes.
03

Using the Asymptotes

For a vertical hyperbola with asymptotes \(y = \pm \frac{a}{b}x\), since \(b = a\), the slope of the asymptotes indicates that \(b/a = 1\). Therefore, \(b = a\). This allows us to set up the equation as \(a = b\).
04

Finding \(c\) from the Foci

The distance \(c\) from the center to a focus of the hyperbola is \(c = \sqrt{2}\). In a hyperbola, \(c^2 = a^2 + b^2\). Substituting \(c = \sqrt{2}\) and \(b = a\), we have:\[ c^2 = a^2 + a^2 = 2a^2 \]Given \(c^2 = 2\), we substitute it in to get:\[ 2 = 2a^2 \]
05

Solving for \(a\) and \(b\)

Solve \(2 = 2a^2\) to find \(a\):\[ a^2 = 1 \]This implies \(a = 1\), and since \(b = a\), \(b = 1\).
06

Final Step: Writing the Equation

Substituting \(a = 1\) and \(b = 1\) into the standard form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the equation becomes:\[ \frac{y^2}{1^2} - \frac{x^2}{1^2} = 1 \]Simplifying, this results in:\[ y^2 - x^2 = 1 \]

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Key Concepts

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

Foci of Hyperbola
The foci of a hyperbola are two fixed points located along the hyperbola's transverse axis. These points are crucial because they help define the curve's shape and orientation. For a hyperbola centered at the origin, the foci are symmetric with respect to the center. In the case of a vertical hyperbola, such as the one given in the exercise, with foci
  • at (0, \pm \sqrt{2}),
we know that the hyperbola opens upwards and downward. The distance from the center to each focus this distance, denoted as \(c\), is \(\sqrt{2}\). This value is instrumental in determining the hyperbola's equation.
Vertices of Hyperbola
The vertices of a hyperbola are points where the hyperbola intersects its transverse axis. They are the nearest points on the hyperbola to the center. Understanding the position of a hyperbola's vertices helps us visualize the hyperbola's orientation and size. For a hyperbola centered at the origin, the vertices' position provides information about the major axis. The distance from the center to a vertex is called \(a\), and it's calculated through the hyperbola's equation. In our case, since the standard form equation is derived as
  • \(y^2 - x^2 = 1\),
we determine \(a = 1\). Thus, the vertices are located at (0, \pm 1). This confirms the hyperbola opens in the vertical direction.
Asymptotes of Hyperbola
Asymptotes are lines that a hyperbola approaches but never touches. These lines guide the overall shape of the hyperbola as it extends infinitely in its opening direction. For a hyperbola centered at the origin, the asymptotes are diagonals that intersect at the center. In the exercise, with given asymptotes
  • as \(y = \pm x\),
we confirm that the hyperbola is centered at the origin and vertically oriented. The slope of these asymptotes allows us to determine that for a vertical hyperbola, the relationship between \(a\) and \(b\) is \(b = a\). This ensures the slope is \(\pm 1\), matching the equations given.
Standard Form of Hyperbola
The standard form equation of a hyperbola describes its geometric properties precisely. For a vertical hyperbola centered at the origin, the equation is
  • \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\).
In this equation:
  • \(a\) represents the semi-distance from the center to a vertex along the transverse axis,
  • \(b\) dictates the semi-distance along the conjugate axis.
To find this equation in the given problem, by using the formula \(c^2 = a^2 + b^2\) with known values, we determined \(a\) and \(b\) to be equal to 1. Substituting these values results in the equation \(y^2 - x^2 = 1\), representing the unique shape and orientation of the hyperbola at the origin with vertical orientation.

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Most popular questions from this chapter

Give equations for ellipses and tell how many units up or down and to the right or left each ellipse is to be shifted. Find an equation for the new ellipse, and find the new foci, vertices, and center. $$\frac{x^{2}}{6}+\frac{y^{2}}{9}=1, \quad \text { left } 2, \text { down } 1$$

The "triangular" region in the first quadrant bounded by the \(x\) -axis, the line \(x=4,\) and the hyperbola \(9 x^{2}-4 y^{2}=36\) is revolved about the \(x\) -axis to generate a solid. Find the volume of the solid.

Give the eccentricities of conic sections with one focus at the origin along with the directrix corresponding to that focus. Find a polar equation for each conic section. $$e=1, \quad x=2$$

Give equations for parabolas and tell how many units up or down and to the right or left each parabola is to be shifted. Find an equation for the new parabola, and find the new vertex, focus, and directrix. $$y^{2}=-12 x, \quad \text { right } 4, \text { up } 3$$

Give equations for parabolas and tell how many units up or down and to the right or left each parabola is to be shifted. Find an equation for the new parabola, and find the new vertex, focus, and directrix. $$y^{2}=4 x, \quad \text { left } 2, \text { down } 3$$
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