/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Find an equation for the hyperbo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find an equation for the hyperbola that satisfies the given conditions. Foci \((0, \pm 1),\) length of transverse axis 1

Short Answer

Expert verified
The equation of the hyperbola is \(y^2 - \frac{1}{3}x^2 = 1\).

Step by step solution

01

Identify the elements of the hyperbola

The given foci are \((0, \pm 1)\). This means the hyperbola is centered at the origin and is vertical. For a vertical hyperbola, the foci are described by the coordinates \((0, \pm c)\), where \(c\) is the distance from the center to each focus. Here, \(c = 1\).
02

Use the transverse axis length to find \(a\)

The length of the transverse axis is 1, meaning \(2a = 1\), where \(a\) is the distance from the center to the vertices along the y-axis. Therefore, \(a = \frac{1}{2}\).
03

Apply the relationship between \(a\), \(b\), and \(c\)

For a hyperbola, the relationship between the semi-major axis \(a\), semi-minor axis \(b\), and the distance to the foci \(c\) is given by the equation \(c^2 = a^2 + b^2\). Since \(c = 1\) and \(a = \frac{1}{2}\), substitute these values to find \(b^2\): \(1^2 = \left(\frac{1}{2}\right)^2 + b^2\).
04

Calculate \(b^2\)

Simplify and solve for \(b^2\): - \(1 = \frac{1}{4} + b^2\).- Subtract \(\frac{1}{4}\) from both sides: - \(b^2 = 1 - \frac{1}{4} = \frac{3}{4}\).
05

Write the equation in standard form

For a vertical hyperbola centered at the origin, the equation is given by:\[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\]Using \(a^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\) and \(b^2 = \frac{3}{4}\), the equation becomes:\[\frac{y^2}{\frac{1}{4}} - \frac{x^2}{\frac{3}{4}} = 1\]
06

Simplify the equation

Simplify the hyperbola's equation by clearing the fractions:Multiply each term by 4 (the least common denominator):\[4 \cdot \frac{y^2}{\frac{1}{4}} - 4 \cdot \frac{x^2}{\frac{3}{4}} = 4 \cdot 1\]This simplifies to:\[4y^2 - \frac{4}{3}x^2 = 4\]Dividing through by 4 to further simplify:\[y^2 - \frac{1}{3}x^2 = 1\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Conic Sections
Conic sections are fascinating curves that are formed by the intersection of a plane and a double-napped cone. This concept encompasses several distinct shapes, including circles, ellipses, parabolas, and hyperbolas. The type of conic section formed depends on the angle at which the plane intersects the cone.

Each conic section has unique equations and properties associated with it:
  • Circles and ellipses are formed when the intersecting plane is perpendicular or at an angle less than the cone's slant height.
  • Hyperbolas, our focus here, occur when the plane cuts through both nappes of the cone.
  • Parabolas result when the plane is parallel to the cone's slant edge.
The beauty of conic sections lies in their wide applicability in fields such as astronomy, physics, and engineering, providing models for everything from planetary motions to the design of reflective surfaces.
Vertical Hyperbola
A vertical hyperbola is one type of hyperbola characterized by its distinct orientation. Unlike its horizontal counterpart, a vertical hyperbola opens upwards and downwards from its center. This orientation is determined by the position of its transverse axis, which runs along the y-axis in this case.

The standard form of a vertical hyperbola centered at the origin is given by the equation:
  • \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)
Here, \(a\) and \(b\) represent the distances from the center to the vertices and to the endpoints of the conjugate axis, respectively. The transverse axis' length, 2a, stretches along the vertical direction, giving the hyperbola its vertical opening.
Understanding the orientation and position of the axes helps in sketching the hyperbola and analyzing its properties, such as symmetry and asymptotes.
Foci and Vertices
In a hyperbola, the foci and vertices are pivotal elements defining its distinct-shaped curves. Both these points help in determining the structure and equation of the hyperbola.

- **Foci:** These are the two fixed points located inside each lobe of the hyperbola. The distances from any point on the hyperbola to each focus differ by a constant. For a vertical hyperbola centered at origin, the foci are located at \((0, \pm c)\), where \(c\) is the distance from the center to each focus.- **Vertices:** These are points where the hyperbola intersects its transverse axis. They are essential as they frame the shape of the graph. For a vertical hyperbola, vertices are found at \((0, \pm a)\), and their separation is the transverse axis' length. The relationship between these components is encapsulated in the equation \(c^2 = a^2 + b^2\), crucial for establishing complete knowledge of the hyperbola's dimensions. This emphasis on foci and vertices also assists in exploring the hyperbola's real-world applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If it is a hyperbola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. $$x^{2}-4 y^{2}-2 x+16 y=20$$

A pair of parametric equations is given. (a) Sketch the curve represented by the parametric equations. (b) Find a rectangular-coordinate equation for the curve by eliminating the parameter. $$x=\cot t, \quad y=\csc t, \quad 0

Use a graphing device to draw the curve represented by the parametric equations. $$x=3 \sin 5 t, \quad y=5 \cos 3 t$$

A Family of Confocal Conics Conics that share a focus are called confocal. Consider the family of conics that have a focus at \((0,1)\) and a vertex at the origin (see the figure). (a) Find equations of two different ellipses that have these properties. (b) Find equations of two different hyperbolas that have these properties. (c) Explain why only one parabola satisfies these properties. Find its equation. (d) Sketch the conics you found in parts (a), (b), and (c) on the same coordinate axes (for the hyperbolas, sketch the top branches only) (e) How are the ellipses and hyperbolas related to the parabola?

(a)Use the discriminant to determine whether the graph of the equation is a parabola, an ellipse, or a hyperbola. (b) Use a rotation of axes to eliminate the \(x y\) -term. (c) Sketch the graph. $$52 x^{2}+72 x y+73 y^{2}=40 x-30 y+75$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.