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Chapter 10: Problem 7

Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. $$y^{2}-\frac{x^{2}}{25}=1$$

Short Answer

Expert verified
Vertices: \((0, 1)\), \((0, -1)\); Foci: \((0, \sqrt{26})\), \((0, -\sqrt{26})\); Asymptotes: \(y = \pm \frac{1}{5} x\).

Step by step solution

01

Recognize the Hyperbola Form

The equation \(y^2 - \frac{x^2}{25} = 1\) is in the standard form of a hyperbola: \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). This indicates a vertical hyperbola.
02

Identify Parameters

In the equation \(y^2 - \frac{x^2}{25} = 1\), compare it with \(\frac{y^2}{1} - \frac{x^2}{25} = 1\). Here, \(a^2 = 1\) implies \(a = 1\), and \(b^2 = 25\) implies \(b = 5\).
03

Calculate the Vertices

For a vertical hyperbola centered at the origin with equation \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the vertices are at \((0, \pm a)\). So the vertices are \((0, 1)\) and \((0, -1)\).
04

Determine the Foci

To find the foci of the hyperbola, use the formula \(c^2 = a^2 + b^2\). Therefore, \(c^2 = 1 + 25 = 26\), which gives \(c = \sqrt{26}\). The foci are \((0, \pm \sqrt{26})\).
05

Find the Asymptotes

The equations of the asymptotes for a vertical hyperbola are \(y = \pm \frac{a}{b} x\). Here, the asymptotes are \(y = \pm \frac{1}{5}x\).
06

Sketch the Graph

Draw the vertical hyperbola centered at origin with vertices at \((0, 1)\) and \((0, -1)\). The foci are at \((0, \sqrt{26})\) and \((0, -\sqrt{26})\). Draw dashed lines for the asymptotes \(y = \frac{1}{5}x\) and \(y = -\frac{1}{5}x\) to complete the sketch.

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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 critical points that help define its shape and orientation. In this particular hyperbola, identified by the equation \(y^2 - \frac{x^2}{25} = 1\), we're dealing with a vertical hyperbola since the \(y^2\) term comes first. For vertical hyperbolas in the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the vertices lie on the y-axis and are located at \((0, \pm a)\).
To find these vertices, we identify \(a^2\) from the equation, which is \(1\). Solving for \(a\), we find \(a = 1\), hence the vertices are \((0, 1)\) and \((0, -1)\).
  • Vertices give you important points to plot initially.
  • The distance between them is directly correlated to the size and steepness of the hyperbola.
These vertices help provide a framework to begin sketching or further analyzing anything involving the hyperbola.
Foci of Hyperbola
The foci of a hyperbola are key points located along the axis of symmetry, further away from the center than the vertices. For the hyperbola \(y^2 - \frac{x^2}{25} = 1\), which is vertical, the foci alignment is along the y-axis. For hyperbolas, the foci are found using the formula \(c^2 = a^2 + b^2\).
In this case, \(a^2 = 1\) and \(b^2 = 25\) which leads us to \(c^2 = 1 + 25 = 26\). Solving for \(c\), we get \(c = \sqrt{26}\). Hence, the foci are located at \((0, \pm \sqrt{26})\).
  • Foci are farther from the center than the vertices and add depth to the hyperbola's symmetry.
  • Distance from the center can affect the "tightness" of the branches.
Having these foci points helps in understanding the degree of "spread" the hyperbola has.
Asymptotes of Hyperbola
Asymptotes are invisible lines that a hyperbola approaches but never touches. They help in defining the broad shape of the hyperbola as they set the directions of its branches. For the vertical hyperbola \(y^2 - \frac{x^2}{25} = 1\), the asymptotes are derived from the slopes represented in \(y = \pm \frac{a}{b} x\).
Here, \(a = 1\) and \(b = 5\), thus the equations become \(y = \pm \frac{1}{5}x\). These asymptotes guide the hyperbola's orientation in space and provide a clear limit on how it behaves as it extends.
  • Asymptotes contribute to the hyperbola's end behavior.
  • They are essential for sketching, especially when curves extend beyond plotted points.
By drawing these lines, you create a framework within which the hyperbola's curve can be precisely sketched or analyzed.

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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. $$4 x^{2}+25 y^{2}-24 x+250 y+561=0$$

Let \(Z, Z^{\prime},\) and \(R\) be the matrices $$\begin{array}{cc}Z=\left[\begin{array}{l}x \\\y\end{array}\right] & Z^{\prime}=\left[\begin{array}{l}X \\ Y\end{array}\right] \\\R=\left[\begin{array}{lr}\cos \phi & -\sin \phi \\\\\sin \phi & \cos \phi\end{array}\right]\end{array}$$ Show that the Rotation of Axes Formulas can be written as $$Z=R Z^{\prime} \quad \text { and } \quad Z^{\prime}=R^{-1} Z$$

(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. $$21 x^{2}+10 \sqrt{3} x y+31 y^{2}=144$$

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. $$9 x^{2}-36 x+4 y^{2}=0$$

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. $$x=2 t, \quad y=t+6$$
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