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Magazine Chapter 10: Problem 22
Graph each hyperbola. Give the domain, range, center, vertices, foci, and equations of the asymptotes for each figure. $$4(x+9)^{2}-25(y+6)^{2}=100$$
Short Answer
Expert verified
Center: (-9, -6). Vertices: (-14, -6) and (-4, -6). Foci: (-9 - sqrt(29), -6) and (-9 + sqrt(29), -6). Asymptotes: y + 6 = 卤(2/5)(x + 9). Domain: (-鈭�, 鈭�). Range: (-鈭�, -11) U (-1, 鈭�).
Step by step solution
01
Rewrite the Equation in Standard Form
Divide each term in the equation by 100 to rewrite it in standard form: \[ \frac{4(x+9)^2}{100} - \frac{25(y+6)^2}{100} = 1 \] which simplifies to \[ \frac{(x+9)^2}{25} - \frac{(y+6)^2}{4} = 1 \]
02
Identify the Center
The standard form of a hyperbola is \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), where \((h,k)\) is the center. For this equation, \(h = -9\) and \(k = -6\). Thus, the center of the hyperbola is \((-9, -6)\).
03
Determine the Vertices
The vertices are found by adding and subtracting \(a\) from the center \((h,k)\). Here, \(a^2 = 25\), so \(a = 5\). The vertices are at \((h \, \pm \, a, k)\), giving us the points \((-9-5, -6)\) and \((-9+5, -6)\). Therefore, the vertices are at \((-14, -6)\) and \((-4, -6)\).
04
Find the Foci
The foci are determined using \(c^2 = a^2 + b^2\). Here, \(a^2 = 25\) and \(b^2 = 4\). So, \(c^2 = 25 + 4 = 29\), thus \(c = \sqrt{29}\). The foci are at \((h \, \pm \, c, k)\), which gives us points \((-9 - \sqrt{29}, -6)\) and \((-9 + \sqrt{29}, -6)\).
05
Find the Asymptotes
The equations of the asymptotes for a hyperbola in standard form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) are given by \[ y - k = \pm \frac{b}{a}(x - h) \] Substituting \(h = -9\), \(k = -6\), \(a = 5\), and \(b = 2\), the equations of the asymptotes are: \[ y + 6 = \pm \frac{2}{5}(x + 9) \]
06
State the Domain and Range
For a hyperbola opening left and right, the domain is all real numbers \((-\infty, \infty)\). The range is all real numbers except between the y-values of the transverse axis at the vertices: \(( -\infty, -6 - 5 ) \cup ( -6 + 5, \infty )\) or simply \((-\infty, -11) \cup (-1, \infty)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hyperbola Standard Form
The standard form of a hyperbola helps us easily identify key features such as its center, vertices, and foci. For the given hyperbola, we start by rewriting the equation in this form. From the equation $$4(x+9)^2 - 25(y+6)^2 = 100$$, we divide every term by 100:
\[ \frac{4(x+9)^2}{100} - \frac{25(y+6)^2}{100} = 1 \]
This simplifies to:
\[ \frac{(x+9)^2}{25} - \frac{(y+6)^2}{4} = 1 \]
Here, we can see that \(a^2 = 25\) and \(b^2 = 4\). So, our standard form is:
\[ \frac{(x+9)^2}{25} - \frac{(y+6)^2}{4} = 1 \]
\[ \frac{4(x+9)^2}{100} - \frac{25(y+6)^2}{100} = 1 \]
This simplifies to:
\[ \frac{(x+9)^2}{25} - \frac{(y+6)^2}{4} = 1 \]
Here, we can see that \(a^2 = 25\) and \(b^2 = 4\). So, our standard form is:
\[ \frac{(x+9)^2}{25} - \frac{(y+6)^2}{4} = 1 \]
Domain and Range
Next, we learn to find the domain and range of the hyperbola. For hyperbolas that open horizontally, the domain encompasses all real numbers, meaning the x-values can be anything. Thus, the domain is:
\[ (-\infty, +\infty) \]
The range, however, is slightly more complex. It excludes the y-values between the vertices. Our vertices have y-coordinates of -6, and a value of 5 extends upwards and downwards. Therefore, the range is:
\[ (-\infty, -11) \cup (-1, \infty) \]
\[ (-\infty, +\infty) \]
The range, however, is slightly more complex. It excludes the y-values between the vertices. Our vertices have y-coordinates of -6, and a value of 5 extends upwards and downwards. Therefore, the range is:
\[ (-\infty, -11) \cup (-1, \infty) \]
Hyperbola Asymptotes
Asymptotes guide the shape of a hyperbola. They are lines the hyperbola approaches but never touches. For our hyperbola in standard form:
\[ \frac{(x+9)^2}{25} - \frac{(y+6)^2}{4} = 1 \]
The equations of the asymptotes are:
\[ y - k = \pm \frac{b}{a}(x - h) \]
Substituting \(h = -9\), \(k = -6\), \(a = 5\), and \(b = 2\):
\[ y + 6 = \pm \frac{2}{5}(x + 9) \]
Resulting in:
\[ y = -6 \pm \frac{2}{5}(x + 9) \]
These asymptotes help in sketching the hyperbola accurately.
\[ \frac{(x+9)^2}{25} - \frac{(y+6)^2}{4} = 1 \]
The equations of the asymptotes are:
\[ y - k = \pm \frac{b}{a}(x - h) \]
Substituting \(h = -9\), \(k = -6\), \(a = 5\), and \(b = 2\):
\[ y + 6 = \pm \frac{2}{5}(x + 9) \]
Resulting in:
\[ y = -6 \pm \frac{2}{5}(x + 9) \]
These asymptotes help in sketching the hyperbola accurately.
Hyperbola Vertices
The vertices are crucial points where each branch of the hyperbola is closest to the center. In standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the vertices are located at \( (h \pm a, k) \). For \(a^2 = 25\), we find \(a = 5\). Our center \((h, k)\) is \((-9, -6)\).
Thus, the vertices are at:
\[ (-9 \pm 5, -6) \]
This gives us points: \((-14, -6)\) and \((-4, -6)\).
Thus, the vertices are at:
\[ (-9 \pm 5, -6) \]
This gives us points: \((-14, -6)\) and \((-4, -6)\).
Hyperbola Foci
Foci are fixed points used in the hyperbola definition. To find them, we use \(c^2 = a^2 + b^2\). For our hyperbola, \(a^2 = 25\) and \(b^2 = 4\). Thus:
\[ c^2 = 25 + 4 = 29 \]
Taking the square root, \(c = \sqrt{29}\).
The foci are located at \( (h \pm c, k) \). With center \((-9, -6)\), they are:
\[ (-9 \pm \sqrt{29}, -6) \]
So, we have points: \((-9 - \sqrt{29}, -6)\) and \((-9 + \sqrt{29}, -6)\).
\[ c^2 = 25 + 4 = 29 \]
Taking the square root, \(c = \sqrt{29}\).
The foci are located at \( (h \pm c, k) \). With center \((-9, -6)\), they are:
\[ (-9 \pm \sqrt{29}, -6) \]
So, we have points: \((-9 - \sqrt{29}, -6)\) and \((-9 + \sqrt{29}, -6)\).
Coordinate Geometry
Coordinate geometry helps us graph and understand hyperbolas in a coordinate plane. By identifying the center, vertices, foci, and asymptotes, we can sketch any hyperbola accurately.
- The center \((-9, -6)\) is the starting point.
- Plot the vertices \((-14, -6)\) and \((-4, -6)\).
- Mark the foci \((-9 - \sqrt{29}, -6)\) and \((-9 + \sqrt{29}, -6)\).
- Sketch the asymptotes \(y + 6 = \pm \frac{2}{5}(x + 9)\).
