Find an equation of the hyperbola that has vertices (+√5,0) and the length of the conjugate axis is 6. Equation: 1
Find an equation of the hyperbola that has vertices (+√5,0) and the length of the conjugate axis is 6. Equation: 1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Problem Statement:**
Find an equation of the hyperbola that has vertices \((\pm \sqrt{5}, 0)\) and the length of the conjugate axis is 6.
**Equation:**
\[ \frac{{\Box}}{1} \]
**Explanation:**
To find the equation of the hyperbola, we start with the standard form of a hyperbola centered at the origin:
\[
\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
\]
**Vertices and Axes:**
- The vertices \((\pm \sqrt{5}, 0)\) indicate that \(a = \sqrt{5}\).
- The length of the conjugate axis is 6, so \(2b = 6\) implying \(b = 3\).
Thus, the equation of the hyperbola becomes:
\[
\frac{x^2}{(\sqrt{5})^2} - \frac{y^2}{3^2} = 1
\]
By simplifying, we get:
\[
\frac{x^2}{5} - \frac{y^2}{9} = 1
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff9a2dab3-72cf-4044-ac86-05d83f420c4a%2Faf954e21-3f62-4a93-9b14-e752f602de8c%2F7owo4x_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find an equation of the hyperbola that has vertices \((\pm \sqrt{5}, 0)\) and the length of the conjugate axis is 6.
**Equation:**
\[ \frac{{\Box}}{1} \]
**Explanation:**
To find the equation of the hyperbola, we start with the standard form of a hyperbola centered at the origin:
\[
\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
\]
**Vertices and Axes:**
- The vertices \((\pm \sqrt{5}, 0)\) indicate that \(a = \sqrt{5}\).
- The length of the conjugate axis is 6, so \(2b = 6\) implying \(b = 3\).
Thus, the equation of the hyperbola becomes:
\[
\frac{x^2}{(\sqrt{5})^2} - \frac{y^2}{3^2} = 1
\]
By simplifying, we get:
\[
\frac{x^2}{5} - \frac{y^2}{9} = 1
\]
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