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Chapter 6: Problem 41

For the following exercises, determine the equation of the hyperbola using the information given. Foci located at \((6,-0),(6,0)\) and eccentricity of 3

Short Answer

Expert verified
The equation of the hyperbola is \(\frac{y^2}{4} - \frac{(x-6)^2}{32} = 1\).

Step by step solution

01

Identify the center and orientation

The foci are located at \((6, -0), (6, 0)\). This means both foci have the same x-coordinate, which indicates that the hyperbola is vertical and the center of the hyperbola is at \((6, 0)\).
02

Find the value of c

In a hyperbola, the distance from the center to each focus, \(c\), is half the distance between the foci. Since the foci share the same x coordinate, the distance is 0 in the y-direction, meaning the distance is 6 - 0 = 6. So, \(c = 6\).
03

Use eccentricity to find a

The eccentricity \(e\) is given as 3. For hyperbolas, \(e = \frac{c}{a}\). We have \(c = 6\) and \(e = 3\), so solve for \(a\):\[ e = \frac{c}{a} \Rightarrow 3 = \frac{6}{a} \Rightarrow a = \frac{6}{3} = 2 \]
04

Find b using the relationship\(b^2 = c^2 - a^2\)

For hyperbolas, \(b^2 = c^2 - a^2\). We have \(c = 6\) and \(a = 2\), so:\[ b^2 = c^2 - a^2 = 6^2 - 2^2 = 36 - 4 = 32\]
05

Write the equation of the hyperbola

Since the hyperbola is vertical and centered at \((h, k) = (6, 0)\), the equation is:\[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\]Substitute \((h, k) = (6, 0)\), \(a^2 = 4\), and \(b^2 = 32\):\[ \frac{(y - 0)^2}{4} - \frac{(x - 6)^2}{32} = 1\]

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Key Concepts

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

Hyperbola Equation
A hyperbola is a type of conic section characterized by its distinct equation. The equation for a hyperbola differs depending on its orientation鈥攈orizontal or vertical. In our case, the hyperbola is vertical, as indicated by the foci having the same x-coordinate.

The general equation for a vertical hyperbola is:
  • \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)
Here:
  • \((h, k)\) represents the center of the hyperbola.
  • \(a\) is the distance from the center to the vertices along the transverse axis.
  • \(b\) is associated with the distance along the conjugate axis.
If a hyperbola is horizontal, swap \(x\) and \(y\) in the equation foundational formula.

The "1" on the right side of the equation is a key feature that cannot be changed when defining a hyperbola in this standard form. It ensures the curve of the conic section does not deviate from that of a standard hyperbola.
Eccentricity
Eccentricity is a fundamental element in determining the shape of a hyperbola. This value, denoted by \(e\), provides insight into how much the conic section deviates from being a circle (where eccentricity would be 0).

For hyperbolas:
  • \( e > 1 \)
Eccentricity is determined by the relationship:
  • \( e = \frac{c}{a} \)
Where:
  • \(c\) is the distance from the center to each focus.
  • \(a\) is the distance from the center to the vertices along the main axis.
In our example, since \(e = 3\), it shows that the foci are significantly farther apart from the center compared to the vertices, indicating a stretched hyperbola. Calculating eccentricity helps not just in sketching hyperbola accurately, but also in understanding its geometric properties.
Center of the Hyperbola
The center of a hyperbola is a vital point that acts as the midpoint between the foci. It serves as the origin for the reference axes used to define the hyperbola's equation.

In our specific problem, the foci are located at
  • \((6, -0)\)
  • \((6, 0)\)
This implies they share the same x-value, indicating the center must be at:
  • \((6, 0)\)
This central point allows you to determine whether the hyperbola is oriented vertically or horizontally. For vertical hyperbolas, the elements like vertices and the transverse axis extend vertically from here. Understanding the center鈥檚 role is crucial in graphing and comprehending the overall symmetry and shape of the hyperbola.

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