Chapter 7: Problem 3
Find the equation of the hyperbola whose eccentricicy is \(\sqrt{2}\) and the distance between the foci is 16 , taking transverse axes and conjugate axis of the hyperbola as \(x\) and \(y\) axes respectively.
Short Answer
Expert verified
The equation of the hyperbola is \( \frac{x^2}{32} - \frac{y^2}{16} = 1 \).
Step by step solution
01
Determine the relationship between the variables
For a hyperbola described by the equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the distance between foci is given by the equation \( 2ae = f \), where \( e \) is the eccentricity, \( a \) the semi-transverse axis and \( f \) is the distance between the foci.
02
Isolate the variable 'a'
From the equation given in step 1, isolate \( a \) to find the semi-transverse axis in terms of the given eccentricity and distance between foci: \( a = \frac{f}{2e} = \frac{16}{2\sqrt{2}} = 4\sqrt{2} \).
03
Use the relationship between semi-transverse axis, semi-conjugate axis, and eccentricity
We know that the equation for the eccentricity of a hyperbola is \( e = \sqrt{1+\frac{b^2}{a^2}} \). Substituting the given eccentricity as well as the calculated value of \( a \) into this equation enables us to isolate and solve for \( b \): \( \sqrt{2} = \sqrt{1+\frac{b^2}{32}} \). Simplifying this equation gives \( b = 4 \).
04
Write the equation of the eccentricity
Having calculated the semi-transverse and semi-conjugate axes, the equation for the hyperbola is \( \frac{x^2}{(4\sqrt{2})^2} - \frac{y^2}{4^2} = 1 \), which simplifies to \( \frac{x^2}{32} - \frac{y^2}{16} = 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Eccentricity in Hyperbolas
Eccentricity, often denoted as \( e \), is a crucial concept in understanding hyperbolas, as well as other conic sections like ellipses and parabolas. In a hyperbola, eccentricity is always greater than 1. It describes how much the conic section diverges from being a circle. The value of eccentricity helps determine the shape and properties of the hyperbola.
In the provided problem, the eccentricity is given as \( \sqrt{2} \). This indicates that the hyperbola is moderately elongated, since its eccentricity is greater than 1 but not too large. The formula for eccentricity for a hyperbola is given as \( e = \sqrt{1 + \frac{b^2}{a^2}} \), where \( a \) is the semi-transverse axis and \( b \) is the semi-conjugate axis.
In the provided problem, the eccentricity is given as \( \sqrt{2} \). This indicates that the hyperbola is moderately elongated, since its eccentricity is greater than 1 but not too large. The formula for eccentricity for a hyperbola is given as \( e = \sqrt{1 + \frac{b^2}{a^2}} \), where \( a \) is the semi-transverse axis and \( b \) is the semi-conjugate axis.
- If \( e = \sqrt{2} \), it provides a key link to find other components of the hyperbola such as the semi-transverse axis \( a \) and semi-conjugate axis \( b \).
- The eccentricity is vital in combining with other known values to form the equation of the hyperbola.
The Role of the Semi-Transverse Axis
The semi-transverse axis, represented as \( a \), is half the length of the transverse axis of a hyperbola. It is a fundamental component in forming the equation of a hyperbola. The semi-transverse axis runs along the major axis and helps in determining the position of the hyperbola's foci and vertices.
Given the problem's parameters, the formula involves \( a \), the semi-transverse axis, in the expression \( 2ae = f \), where \( f \) is the distance between the foci. From this, we can calculate \( a \) using:
\( a = \frac{f}{2e} = \frac{16}{2\sqrt{2}} = 4\sqrt{2} \).
Given the problem's parameters, the formula involves \( a \), the semi-transverse axis, in the expression \( 2ae = f \), where \( f \) is the distance between the foci. From this, we can calculate \( a \) using:
\( a = \frac{f}{2e} = \frac{16}{2\sqrt{2}} = 4\sqrt{2} \).
- This value of \( a \) is associated with the directrix, a line related to the hyperbola, which defines how much the hyperbola opens.
- The length of the semi-transverse axis is critical to understanding the hyperbola's width and the calculation of its other axis, the semi-conjugate axis.
Calculating the Distance Between Foci
The distance between the foci of a hyperbola, represented as \( f \), is crucial for constructing the hyperbola's equation. It directly involves the concept of eccentricity and the semi-transverse axis. The formula \( 2ae = f \) connects these elements, where \( 2a \) is the length of the transverse axis, \( e \) is the eccentricity, and \( f \) is the total distance between the two foci.
For this specific problem, the distance is given as 16. By substituting \( f = 16 \) and \( e = \sqrt{2} \) into the formula, you can solve for \( a \). This provides the necessary measure of the hyperbola's transverse axis to use in its general equation.
For this specific problem, the distance is given as 16. By substituting \( f = 16 \) and \( e = \sqrt{2} \) into the formula, you can solve for \( a \). This provides the necessary measure of the hyperbola's transverse axis to use in its general equation.
- The foci, being a fixed distance apart, further characterize the hyperbola's openness and orientation.
- Understanding the role of the distance between foci aids in comprehending how the hyperbola encases these focal points as it expands in both directions along the transverse axis.