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Chapter 10: Problem 29

In Exercises \(23-30\) , find the eccentricity of the hyperbola. Then find and graph the hyperbola's foci and directrices. $$ 8 y^{2}-2 x^{2}=16 $$

Short Answer

Expert verified
Eccentricity is \( \sqrt{5} \); Foci: \((0, \pm \sqrt{10})\); Directrices: \(y = \pm \frac{\sqrt{10}}{5}\).

Step by step solution

01

Simplify the equation

The given equation is \( 8y^2 - 2x^2 = 16 \). Divide every term by 16 to simplify it into standard form. The equation becomes \( \frac{y^2}{2} - \frac{x^2}{8} = 1 \). This represents a hyperbola centered at the origin with the transverse axis along the y-axis.
02

Identify hyperbola parameters

From the standard form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), identify \( a^2 = 2 \) and \( b^2 = 8 \). Thus, \( a = \sqrt{2} \) and \( b = \sqrt{8} = 2\sqrt{2} \).
03

Calculate the eccentricity

For hyperbolas, the eccentricity \( e \) is calculated as \( e = \sqrt{1 + \frac{b^2}{a^2}} \). Substitute the values to get \( e = \sqrt{1 + \frac{8}{2}} = \sqrt{5} \). The eccentricity is \( e = \sqrt{5} \).
04

Find the foci

The coordinates for the foci of a hyperbola are \( (0, \pm c) \) for a hyperbola opening vertically. Find \( c \) using \( c = ae = \sqrt{2} \times \sqrt{5} = \sqrt{10} \). Thus, the foci are at \( (0, \pm \sqrt{10}) \).
05

Determine and graph the directrices

The equations for the directrices are \( y = \pm \frac{a^2}{c} \). Calculate \( \frac{a^2}{c} = \frac{2}{\sqrt{10}} = \frac{\sqrt{10}}{5} \) after rationalizing the denominator. Thus, the directrices are \( y = \pm \frac{\sqrt{10}}{5} \). To graph, plot the foci and draw the lines for the directrices parallel to the x-axis at these y-values.

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Key Concepts

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

Eccentricity
In hyperbolas, eccentricity measures how "stretched" the conic section is. It is a crucial parameter that helps in distinguishing hyperbolas from other conic sections like ellipses or circles. Eccentricity, denoted as \( e \), of a hyperbola is always greater than 1. In this problem, it is calculated using the formula:
  • \( e = \sqrt{1 + \frac{b^2}{a^2}} \)
Here, \( a \) and \( b \) are derived from the equation \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).
For our specific hyperbola with \( a^2 = 2 \) and \( b^2 = 8 \), substituting these values gives \( e = \sqrt{5} \).
This shows that the hyperbola is significantly elongated along its transverse axis.
Foci
The foci of a hyperbola are points that lie on the transverse axis and are symmetrical about the center of the hyperbola. For a hyperbola oriented along the y-axis, like the one given, the foci have coordinates \( (0, \pm c) \). The value of \( c \) defines how far apart these foci are from the origin.
  • Calculate \( c \) as \( c = ae = \sqrt{2} \times \sqrt{5} = \sqrt{10} \)
  • The foci are \( (0, \pm \sqrt{10}) \)
These foci are crucial in defining the hyperbola's geometric properties and are not located on the curve itself but help form it.
Each point on the hyperbola maintains a constant difference in distance from each focus.
Directrices
The directrices of a hyperbola are lines parallel to the conjugate axis that help define its shape. For the equation \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the directrices are given by the equations:
  • \( y = \pm \frac{a^2}{c} \)
In this scenario, after computing \( c = \sqrt{10} \) and \( a^2 = 2 \), the directrices are found by:
  • \( y = \pm \frac{2}{\sqrt{10}} = \pm \frac{\sqrt{10}}{5} \)
The directrices help in constructing the hyperbola's asymptotes and are an extension of its geometric property.
Graphing Hyperbolas
Graphing a hyperbola requires plotting several components: the center, the foci, the vertices, and the directrices. Given the equation \( \frac{y^2}{2} - \frac{x^2}{8} = 1 \), the graph centers at the origin. Here, the transverse axis is along the y-axis.
To accurately portray the hyperbola:
  • Identify the vertices at \( (0, \pm \sqrt{2}) \).
  • Plot the foci points \( (0, \pm \sqrt{10}) \).
  • Draw the directrices at \( y = \pm \frac{\sqrt{10}}{5} \).
By connecting these points and considering the asymptotic behavior (lines through opposite vertices of the rectangle formed by \( a \) and \( b \)), a complete and accurate graph of the hyperbola is created.
This visual representation helps in better understanding the structure and properties of the hyperbola.

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Most popular questions from this chapter

Find the areas of the regions Inside the lemniscate \(r^{2}=6 \cos 2 \theta\) and outside the circle \(r=\sqrt{3}\)

Exercises \(29-36\) 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. $$ e=1 / 5, \quad y=-10 $$

Exercises \(49-52\) 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. $$ \frac{x^{2}}{16}+\frac{y^{2}}{25}=1, \quad \text { left } 4, \text { down } 5 $$

Which of the following has the same graph as \(r=1-\cos \theta ?\) $$ \text { a. }r=-1-\cos \theta \quad \text { b. } r=1+\cos \theta $$ Confirm your answer with algebra.

In Exercises \(65-68\) , find a polar equation for the given curve. In each case, sketch a typical curve. $$ x \cos \alpha+y \sin \alpha=p \quad(\alpha, p \text { constant }) $$
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