Chapter 8: Problem 56
Write an equation for each conic. Each parabola has vertex at the origin, and each ellipse or hyperbola is centered at the origin. $$\text { Focus }(2,0) ; e=\frac{6}{5}$$
Short Answer
Expert verified
The equation of the hyperbola is \( \frac{9x^2}{25} - \frac{9y^2}{11} = 1 \).
Step by step solution
01
Identify the Conic Type
The given eccentricity \( e = \frac{6}{5} \) is greater than \( 1 \), which indicates that the conic section is a hyperbola.
02
Determine the Orientation
Since the focus is at \((2,0)\), the transverse axis is horizontal. Therefore, the hyperbola will have the equation of the form: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
03
Use the Eccentricity Formula
For a hyperbola, the eccentricity \( e \) is defined as \( e = \frac{c}{a} \), where \( c \) is the distance from the center to the focus, \( c = 2 \). So, \( \frac{6}{5} = \frac{2}{a} \).
04
Solve for \( a \)
From \( \frac{6}{5} = \frac{2}{a} \), solve for \( a \):\[ a = \frac{2 imes 5}{6} = \frac{10}{6} = \frac{5}{3} \].
05
Calculate \( b^2 \) using Hyperbola Formula
Recall that for each hyperbola \( c^2 = a^2 + b^2 \). We have \( a = \frac{5}{3} \), so \( a^2 = \left(\frac{5}{3}\right)^2 = \frac{25}{9} \) and \( c^2 = 2^2 = 4 \).Substitute into the equation:\[ 4 = \frac{25}{9} + b^2 \].
06
Solve for \( b^2 \)
Rearrange the equation to solve for \( b^2 \):\[ b^2 = 4 - \frac{25}{9} = \frac{36}{9} - \frac{25}{9} = \frac{11}{9} \].
07
Write the Equation of the Hyperbola
With \( a^2 = \frac{25}{9} \) and \( b^2 = \frac{11}{9} \), the standard form of the equation is:\[ \frac{x^2}{\frac{25}{9}} - \frac{y^2}{\frac{11}{9}} = 1 \],which simplifies to:\[ \frac{9x^2}{25} - \frac{9y^2}{11} = 1 \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hyperbola
A hyperbola is a type of conic section that is formed by the intersection of a plane with both nappes of a cone. Unlike an ellipse, a hyperbola opens outward in two opposite directions. A simple way to remember its shape is to think of it as two mirrored curves resembling open-ended U shapes.
- A hyperbola has two branches, each resembling half of a parabola but with more symmetry.
- It is defined by two focal points, and the difference in the distances from these focal points to any point on the hyperbola is constant.
- The standard form of a hyperbola centered at the origin depends on its orientation.
Eccentricity
Eccentricity, denoted as \( e \), is a parameter that defines the shape of a conic section. It is a measure of how much a conic deviates from being circular.
- For a hyperbola, the eccentricity is always greater than 1.
- This parameter helps distinguish conic sections: \( e = 0 \) for circles, \( 0 < e < 1 \) for ellipses, \( e = 1 \) for parabolas, and \( e > 1 \) for hyperbolas.
- An eccentricity greater than one indicates a hyperbola and suggests that the curves open wide.
- The given example with \( e = \frac{6}{5} \) confirms that the conic is indeed a hyperbola as \( \frac{6}{5} > 1 \).
Transverse Axis
The transverse axis of a hyperbola is a key feature that defines its orientation and shape. It is the axis that passes through both foci and is the line segment between the vertices.
- In a horizontal hyperbola, the transverse axis lies along the x-axis.
- In a vertical hyperbola, it lies along the y-axis.
- This axis provides a fundamental basis for determining the equation of the hyperbola.