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Chapter 13: Problem 31

Graph each hyperbola. See Example 3. $$ \frac{(y-3)^{2}}{25}-\frac{x^{2}}{25}=1 $$

Short Answer

Expert verified
To graph the hyperbola, plot the center at (0, 3), vertices at (0, 8) and (0, -2), and sketch asymptotes with equations y = 3 卤 x.

Step by step solution

01

Identify the form of the hyperbola

The given equation \( \frac{(y-3)^{2}}{25}-\frac{x^{2}}{25}=1 \) is in the standard form of a hyperbola \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \). This indicates a vertical hyperbola with the center at \((h, k)\).
02

Determine the center of the hyperbola

From the equation, we identify \( h = 0 \) and \( k = 3 \). Thus, the center of the hyperbola is \((0, 3)\).
03

Identify the transverse and conjugate axes

For vertical hyperbolas, the transverse axis is vertical. The length of the semi-major axis (transverse axis) is given by \(a\). From our equation, \(a^2 = 25\) which gives \(a = 5\). The conjugate axis, which is horizontal here, has a length determined by \(b\). Here, \(b^2 = 25\) which gives \(b = 5\).
04

Plot the vertices

The vertices occur along the transverse axis, symmetrically from the center. With \(a = 5\), the vertices are at \((0, 3 + 5) = (0, 8)\) and \((0, 3 - 5) = (0, -2)\). These points are marked as vertices of the hyperbola.
05

Sketch the asymptotes

For vertical hyperbolas, the equations of the asymptotes can be found as \( y = k \pm \frac{a}{b}x \). In our case this becomes \( y = 3 \pm x \). Draw these lines through the center as they provide guidance for sketching the hyperbola.

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Key Concepts

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

Standard Form of Hyperbola
The equation \( \frac{(y-3)^{2}}{25}-\frac{x^{2}}{25}=1 \) is in the standard form of a hyperbola. This form is written as \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \) for a vertical hyperbola. Here's how to recognize it:
  • The template \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \) indicates a hyperbola where the \( y \)-variable is first.
  • A negative sign between the terms denotes a hyperbola.
  • If \( y \) comes first, the hyperbola opens along the \( y \)-axis.
In this equation, since \( k = 3 \) and \( h = 0 \), the center of the hyperbola is located at \((0, 3)\). Identifying the center is crucial because it's the reference point for plotting other components like vertices and asymptotes.
Vertices of Hyperbola
Vertices are key points where the hyperbola intersects its transverse axis. For our equation, the transverse axis is vertical, thanks to the form \( \frac{(y-k)^{2}}{a^{2}} \). Here's how the calculation works:
  • Identify the length \( a \) which is \( \sqrt{25} = 5 \).
  • Since the transverse axis is vertical, add and subtract \( a \) from the \( y \)-value of the center to find vertices.
  • The center is \((0, 3)\), so the vertices are \((0, 3+5)\) and \((0, 3-5)\).
So, our vertices are \((0, 8)\) and \((0, -2)\). These points define the extent of the hyperbola along the transverse axis.
Asymptotes of Hyperbola
Asymptotes are essential guides that show how the branches of the hyperbola behave at infinity. They pass through the center and provide direction:
  • For vertical hyperbolas, the asymptotes can be calculated with \( y = k \pm \frac{a}{b}x \).
  • Here, \( a = 5 \) and \( b = 5 \), giving us \( y = 3 \pm x \).
This results in two asymptote lines: \( y = 3 + x \) and \( y = 3 - x \). Visually, they help bracket the curve of the hyperbola, even though the hyperbola never actually touches these lines.
Transverse and Conjugate Axes
Understanding these axes is crucial for grasping the geometry of a hyperbola. They define the orientation:
  • The transverse axis is the axis along which the hyperbola opens. In a vertical hyperbola, it is vertical.
  • For our hyperbola, this axis has a length of \( 2a \) or \( 10 \), because \( a = 5 \).
  • The conjugate axis is perpendicular to the transverse axis. Here, it is horizontal with length \( 2b \) or \( 10 \), since \( b = 5 \).
These axes create a rectangle that aids in drawing the hyperbola. The ellipse-like enclosure, formed by extending the conjugate and transverse axes, provides a framework for sketching the curves accurately.

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Most popular questions from this chapter

$$ \text { Solve the system for real solutions: }\left\\{\begin{array}{l} \frac{1}{x}+\frac{2}{y}=1 \\ \frac{2}{x}-\frac{1}{y}=\frac{1}{3} \end{array}\right. $$

Find the center and radius of each circle. A. \((x-4)^{2}+(y+7)^{2}=28\) B. \((x+4)^{2}+(y-7)^{2}=28\)

Solve each system of equations for real values of x and y. $$ \left\\{\begin{array}{l} x^{2}-y^{2}=4 \\ x+y=4 \end{array}\right. $$

Solve. $$ \left|\frac{4-3 x}{5}\right|=12 $$

$$ \text { Solve the system for real solutions: }\left\\{\begin{array}{l} \frac{1}{x}+\frac{3}{y}=4 \\ \frac{2}{x}-\frac{1}{y}=7 \end{array}\right. $$
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