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Chapter 11: Problem 68

Analyze each equation. \(\frac{(y+2)^{2}}{16}-\frac{(x-2)^{2}}{4}=1\)

Short Answer

Expert verified
This equation represents a hyperbola with center (2, -2), vertices (2, 2) and (2, -6), and foci at (2, -2 \pm 2\sqrt{5})

Step by step solution

01

Identify the Type of Conic Section

The given equation is in the form \(\frac{(y+2)^{2}}{16}-\frac{(x-2)^{2}}{4}=1\). This is the standard form of a hyperbola.
02

Determine the Transverse and Conjugate Axes

For the hyperbola \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), the transverse axis is vertical because the \(y\) term is positive. Here \(a^2 = 16\) and \(b^2 = 4 \), making \(a = 4\) and \(b = 2\).
03

Identify the Center of the Hyperbola

The center \( (h, k) \) of the hyperbola is determined by the values inside the parentheses: \( h = 2 \) and \( k = -2 \). Hence, the center is \( (2, -2) \).
04

Determine the Vertices

The vertices of the hyperbola are found \( \pm a \) units from the center along the transverse axis. This means \( (2, -2 \pm 4) \) resulting in vertices at \( (2, 2) \) and \( (2, -6) \).
05

Determine the Foci

The foci are determined using the equation \(c^2 = a^2 + b^2 \). Here, \( a^2 = 16 \) and \( b^2 = 4\), so \( c^2 = 16 + 4 = 20 \) leading to \( c = \sqrt{20} = 2\sqrt{5}\). The foci are located at \( (2, -2 \pm 2\sqrt{5}) \).

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Key Concepts

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

Conic Sections
Conic sections are the curves obtained by intersecting a plane with a double-napped cone. The different types of conic sections include circles, ellipses, parabolas, and hyperbolas. Each conic section has its distinct properties and equations. A hyperbola, like the one given in the exercise \(\frac{(y+2)^{2}}{16}-\frac{(x-2)^{2}}{4}=1\), is characterized by two separate curves called branches. Understanding conic sections is fundamental as they appear in various real-world applications, bridging concepts from geometry and algebra.
Hyperbola Properties
A hyperbola is a type of conic section defined by its two distinct branches that curve away from each other. The basic properties of a hyperbola include:
  • The standard form equation: \(\frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}}=1\).
  • Two asymptotes that intersect at the center of the hyperbola.
  • The transverse axis and conjugate axis which help to define its shape.
In the given equation, \(\frac{(y+2)^{2}}{16}-\frac{(x-2)^{2}}{4}=1\), the positive term is associated with \(y\). This implies the branches open upwards and downwards, making the transverse axis vertical. Recognizing these properties helps us to graph and further analyze the hyperbola's structure.
Transverse Axis
The transverse axis of a hyperbola is the line segment that passes through both of its vertices. It lies along the direction in which the two branches of the hyperbola open. In the equation \(\frac{(y+2)^{2}}{16}-\frac{(x-2)^{2}}{4}=1\), the transverse axis is vertical. We know this because the term with \(y\) is positive. For this hyperbola:
  • \(a^{2}=16\) which gives \(a=4\).
  • The transverse axis reaches up and down from the center.
Calculating the vertices involves moving a distance 'a' (4 units) up and down from the center, leading to vertices at (2, 2) and (2, -6). Understanding the transverse axis is crucial for visualizing the orientation and shape of the hyperbola.
Vertices and Foci of Hyperbola
Vertices and foci are key features of a hyperbola.
  • The vertices are the closest points to the center on each branch.
  • The foci, important for the hyperbola's definition, lie along the transverse axis further from the vertices.
To find the vertices:
Start from the center (2, -2) and move vertically by 'a' (4 units), getting vertices at (2, 2) and (2, -6).
To find the foci:
Use the relationship \(c^{2} = a^{2} + b^{2}\). Here, \(a^{2} = 16\) and \(b^{2} = 4\), so \(c^{2} = 20\). Thus \(c = \sqrt{20} = 2\sqrt{5} \).
The foci are at (2, -2 \pm \ 2\sqrt{5}) which pinpoint their locations relative to the center of the hyperbola. Having a strong grasp of how to locate the vertices and foci aids significantly in understanding and drawing hyperbolas.

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