Chapter 6: Problem 9
Put the equation \(9 x^{2}-16 y^{2}+36 x+32 y-124=0\) into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?
Short Answer
Expert verified
The hyperbola's asymptotes are \(y = 1 \pm \frac{3}{4}(x + 2)\).
Step by step solution
01
Rearrange the Equation
Start by grouping the terms in the equation by variables: \(9x^2 + 36x - 16y^2 + 32y = 124\).
02
Complete the Square for x
Take the x-terms, \(9x^2 + 36x\). Factor out the 9: \(9(x^2 + 4x)\).To complete the square for \(x^2 + 4x\), add and subtract 4 inside the parenthesis: \(9((x+2)^2 - 4)\), which simplifies to \(9(x+2)^2 - 36\).
03
Complete the Square for y
Take the y-terms, \(-16y^2 + 32y\). Factor out the -16: \(-16(y^2 - 2y)\).To complete the square for \(y^2 - 2y\), add and subtract 1 inside the parenthesis: \(-16((y-1)^2 - 1)\), which simplifies to \(-16(y-1)^2 + 16\).
04
Form the Standard Equation
Substitute the completed square expressions back into the equation:\[9(x+2)^2 - 36 - 16(y-1)^2 + 16 = 124\]Combine constant terms and simplify:\[9(x+2)^2 - 16(y-1)^2 = 144\]Divide the entire equation by 144 to obtain the standard form:\[\frac{(x+2)^2}{16} - \frac{(y-1)^2}{9} = 1\]
05
Identify the Asymptotes
The standard form \(\frac{(x+2)^2}{16} - \frac{(y-1)^2}{9} = 1\) suggests a horizontal hyperbola with center \((-2, 1)\).The asymptotes of such a hyperbola are given by:\(y - 1 = \pm \frac{3}{4}(x + 2)\).Thus, the equations of the asymptotes are:\[y = 1 \pm \frac{3}{4}(x + 2)\].
06
Graph the Hyperbola
Plot the center of the hyperbola at \((-2, 1)\). Draw the asymptotes using the equations \(y = 1 + \frac{3}{4}(x + 2)\) and \(y = 1 - \frac{3}{4}(x + 2)\).Sketch the branches of the hyperbola, ensuring they approach the asymptotes, centered around \((-2, 1)\) and with vertices along the x-axis at \((-6, 1)\) and \((2, 1)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
Completing the square is a method used to transform a quadratic expression into a perfect square trinomial. This process is crucial for reformatting hyperbola equations into their standard form.
In the exercise, you start with the x-terms of the equation, which is initially written as \(9x^2 + 36x\). First, you factor out the 9 from both terms, resulting in \(9(x^2 + 4x)\).
Next, you complete the square for \(x^2 + 4x\) by adding and subtracting 4 within the parentheses. This transforms the expression into a perfect square trinomial: \((x+2)^2\). Consequently, it becomes \(9((x+2)^2 - 4)\) after simplification. Similarly, for the y-terms \(-16y^2 + 32y\), this process yields \(-16((y-1)^2 - 1)\) after you factor out -16 and complete the square by adjusting inside the parentheses.
The complete square technique simplifies algebraic manipulation and sets the expressions up for use in standard form equations, which is very useful for analyzing hyperbolas.
In the exercise, you start with the x-terms of the equation, which is initially written as \(9x^2 + 36x\). First, you factor out the 9 from both terms, resulting in \(9(x^2 + 4x)\).
Next, you complete the square for \(x^2 + 4x\) by adding and subtracting 4 within the parentheses. This transforms the expression into a perfect square trinomial: \((x+2)^2\). Consequently, it becomes \(9((x+2)^2 - 4)\) after simplification. Similarly, for the y-terms \(-16y^2 + 32y\), this process yields \(-16((y-1)^2 - 1)\) after you factor out -16 and complete the square by adjusting inside the parentheses.
The complete square technique simplifies algebraic manipulation and sets the expressions up for use in standard form equations, which is very useful for analyzing hyperbolas.
Standard Form of Hyperbola
The standard form of a hyperbola is a key representation that makes it easier to identify its key characteristics like center, vertices, and asymptotes. Once completing the square for both x and y parts in their respective sections, these expressions are substituted back into the rewritten equation.
The goal is to achieve a standard form that commonly looks like \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), where \((h, k)\) denotes the hyperbola's center.
For our exercise, after substituting the completed squares back and simplifying the equation, we arrive at \[\frac{(x+2)^2}{16} - \frac{(y-1)^2}{9} = 1\]. This implies the hyperbola is horizontally oriented with its center at \((-2, 1)\), with \(a^2 = 16\) and \(b^2 = 9\). By placing the equation into standard form, you can easily extract these properties.
This format is fundamental since it allows direct application of the hyperbola properties and facilitates graphing by clearly indicating the directional orientation and scale.
The goal is to achieve a standard form that commonly looks like \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), where \((h, k)\) denotes the hyperbola's center.
For our exercise, after substituting the completed squares back and simplifying the equation, we arrive at \[\frac{(x+2)^2}{16} - \frac{(y-1)^2}{9} = 1\]. This implies the hyperbola is horizontally oriented with its center at \((-2, 1)\), with \(a^2 = 16\) and \(b^2 = 9\). By placing the equation into standard form, you can easily extract these properties.
This format is fundamental since it allows direct application of the hyperbola properties and facilitates graphing by clearly indicating the directional orientation and scale.
Asymptotes of Hyperbola
Hyperbolas are famous for having asymptotes—lines that the hyperbola curves never actually touch but get infinitely close to. When a hyperbola is in its standard form, the asymptotes can be determined easily.
The standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) suggests that the equations of the asymptotes would be determined by the slope \(\pm \frac{b}{a}\). For horizontal hyperbolas like the one in this exercise, the formula for their asymptotes extends to \(y = k \pm \frac{b}{a}(x - h)\).
Applying this to our example: given that \(h = -2, k = 1, a = 4,\) and \(b = 3\), thus \(\frac{b}{a} = \frac{3}{4}\), the asymptote equations become \[y = 1 \pm \frac{3}{4}(x + 2)\].
These equations guide you when sketching the hyperbola, showing where the branches will approach — yet never intersect — the asymptotes. Understanding and identifying them is pivotal for comprehending hyperbolic structure and behavior.
The standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) suggests that the equations of the asymptotes would be determined by the slope \(\pm \frac{b}{a}\). For horizontal hyperbolas like the one in this exercise, the formula for their asymptotes extends to \(y = k \pm \frac{b}{a}(x - h)\).
Applying this to our example: given that \(h = -2, k = 1, a = 4,\) and \(b = 3\), thus \(\frac{b}{a} = \frac{3}{4}\), the asymptote equations become \[y = 1 \pm \frac{3}{4}(x + 2)\].
These equations guide you when sketching the hyperbola, showing where the branches will approach — yet never intersect — the asymptotes. Understanding and identifying them is pivotal for comprehending hyperbolic structure and behavior.
