Chapter 10: Problem 17
Sketch the graph of the given equation. $$ \frac{(x+3)^{2}}{4}-\frac{(y+2)^{2}}{16}=1 $$
Short Answer
Expert verified
The hyperbola has a center at (-3, -2) and vertices at (-5, -2) and (-1, -2).
Step by step solution
01
Identify the Equation Type
The given equation \( \frac{(x+3)^{2}}{4}-\frac{(y+2)^{2}}{16}=1 \) is in the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), which represents a hyperbola. This specific form indicates a hyperbola oriented horizontally.
02
Determine the Center
The center of the hyperbola can be found from the values \( h \) and \( k \) in the standard form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). For this equation, \( h = -3 \) and \( k = -2 \). Thus, the center is \((-3, -2)\).
03
Identify the Axes Lengths
The lengths of the axes are determined by the denominators in the standard form. For \( \frac{(x+3)^2}{4} \), \( a^2 = 4 \) which gives \( a = 2 \). For \( \frac{(y+2)^2}{16} \), \( b^2 = 16 \) which gives \( b = 4 \). The transverse axis is horizontal and has a length of \( 2a = 4 \), while the conjugate axis is vertical with a length of \( 2b = 8 \).
04
Plot the Center
On a coordinate plane, mark the center of the hyperbola at the point \((-3, -2)\).
05
Draw the Transverse and Conjugate Axes
From the center, move \( a = 2 \) units to the left and right along the horizontal direction to determine points \((-5, -2)\) and \((-1, -2)\). Then, move \( b = 4 \) units up and down vertically to establish points \((-3, 2)\) and \((-3, -6)\). These lines help in sketching the hyperbola.
06
Sketch the Asymptotes
The asymptotes of a hyperbola with the standard form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) are found using \( y-k = \pm \frac{b}{a}(x-h) \). Substituting \( h = -3, k = -2, a = 2, b = 4 \), the equations of the asymptotes are \( y + 2 = \pm 2(x + 3) \). Plot these lines to form diagonals of the rectangle centered at \((-3, -2)\).
07
Sketch the Hyperbola
Using the center, axes lengths, and asymptotes, sketch the hyperbola. It comprises two branches opening to the left and right. The vertices \((-5, -2)\) and \((-1, -2)\) indicate where each branch turns, extending towards the asymptotes.
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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 curves obtained by intersecting a plane with a double cone. Depending on the angle and position of the intersection, different shapes are formed: circles, ellipses, parabolas, and hyperbolas.
For example:
For example:
- Circles and ellipses are formed when the plane cuts across the cone at an angle less than perpendicular.
- Parabolas occur when the plane is parallel to a generating line of the cone.
- Hyperbolas arise when the plane intersects both halves of the double cone.
Standard Form of Hyperbola
The standard form of a hyperbola helps identify its key characteristics and direction. For a horizontally oriented hyperbola, the equation is given by: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]Here,
This formula helps in accurately plotting and understanding the shape and direction of the hyperbola on a graph.
- \((h, k)\) is the center of the hyperbola.
- \(a\) is the distance from the center to each vertex on the transverse axis.
- \(b\) is the distance along the conjugate axis.
This formula helps in accurately plotting and understanding the shape and direction of the hyperbola on a graph.
Axes of Symmetry
The axes of symmetry in a hyperbola significantly contribute to its shape. These axes help to maintain the mirror-image nature of the hyperbola branches on a graph.
For a horizontally oriented hyperbola like the one we're dealing with:
For a horizontally oriented hyperbola like the one we're dealing with:
- The transverse axis runs horizontally and passes through the center, having its endpoints as vertices. It's length is \(2a\).
- The conjugate axis runs vertically, perpendicular to the transverse axis, and is defined by \(2b\).
Asymptotes of Hyperbola
Asymptotes give a hyperbola its characteristic open shape by guiding its branches towards infinity. They are imaginary lines that the branches approach but never touch.
For a hyperbola in the standard form:\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]The asymptotes are given by the equations:\[ y-k = \pm \frac{b}{a}(x-h) \]For the hyperbola with equation \( \frac{(x+3)^2}{4} - \frac{(y+2)^2}{16} = 1 \), and using the values \( h = -3 \), \( k = -2 \), \( a = 2 \), \( b = 4 \), the asymptotes are \( y + 2 = \pm 2(x + 3) \).
These lines cross at the center of the hyperbola \((-3, -2)\) and converge diagonally, creating a bounding region where the hyperbola's branches spread outwards.
For a hyperbola in the standard form:\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]The asymptotes are given by the equations:\[ y-k = \pm \frac{b}{a}(x-h) \]For the hyperbola with equation \( \frac{(x+3)^2}{4} - \frac{(y+2)^2}{16} = 1 \), and using the values \( h = -3 \), \( k = -2 \), \( a = 2 \), \( b = 4 \), the asymptotes are \( y + 2 = \pm 2(x + 3) \).
These lines cross at the center of the hyperbola \((-3, -2)\) and converge diagonally, creating a bounding region where the hyperbola's branches spread outwards.