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Magazine Chapter 7: Problem 5
Graph the hyperbola. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes. $$ \frac{(x+4)^{2}}{16}-\frac{(y-4)^{2}}{1}=1 $$
Short Answer
Expert verified
The center is (-4, 4). Vertices: (-8, 4) and (0, 4). Foci: (-4卤鈭�17, 4). Asymptotes: y = 4卤1/4(x+4).
Step by step solution
01
Rewrite in Standard Form
Identify the standard form of the hyperbola equation. The given equation is \( \frac{(x+4)^2}{16} - \frac{(y-4)^2}{1} = 1 \), which is similar to \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). This confirms it's a horizontal hyperbola.
02
Identify the Center
The center \((h, k)\) of the hyperbola is derived from \((x + 4)^2\) and \((y - 4)^2\), giving \(h = -4\) and \(k = 4\). Thus, the center is \((-4, 4)\).
03
Determine the Axes
For a horizontal hyperbola, the transverse axis is a horizontal line passing through \(y = k\), or \(y = 4\). The conjugate axis is vertical, passing through \(x = h\), or \(x = -4\).
04
Calculate the Vertices
The vertices are located \(a\) units from the center along the transverse axis. Here, \(a^2 = 16\) so \(a = 4\). Thus, the vertices are \((-4 \, \pm \, 4, 4)\), or \((-8, 4)\) and \((0, 4)\).
05
Find the Foci
The foci are calculated using \(c^2 = a^2 + b^2\). Here, \(b^2 = 1\), so \(c^2 = 16 + 1 = 17\), and \(c = \sqrt{17}\). The foci are \((-4 \, \pm \, \sqrt{17}, 4)\).
06
Derive Equations of Asymptotes
The asymptotes for a horizontal hyperbola are given by \(y = k \pm \frac{b}{a}(x - h)\). Thus, the equations are \(y = 4 \pm \frac{1}{4}(x + 4)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equation of a Hyperbola
In mathematics, a hyperbola is a type of conic section that features two symmetrical branches. The equation of a hyperbola is from a set equation format, which can either be horizontal or vertical depending on its opening direction. The standard form for a horizontal hyperbola is \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \].Here:
- \( (h, k) \) is the center of the hyperbola.
- \( a^2 \) and \( b^2 \) are the squares of the distances from the center to the vertices and the distances to the co-vertices, respectively.
Graphing Hyperbolas
Graphing hyperbolas might seem tricky at first, but it becomes straightforward once you break it down. Follow these steps:
- Identify the center: Use the equations inside the squared terms to locate the center point \((h, k)\).
- Draw the transverse axis: For horizontal hyperbolas, this lies along the x-axis, whereas for vertical hyperbolas, it falls along the y-axis.
- Plot the vertices: Measure \(a\) units each direction along the transverse axis from the center.
- Plot the foci: Use the equation \(c^2 = a^2 + b^2\) to find \(c\) and measure from the center along the transverse axis.
- Draw the asymptotes: These are lines that the hyperbola approaches but never touches.
Hyperbola Center
The center of a hyperbola is akin to the heart of its graph. It's the midpoint of the conjugate and transverse axes and serves as a positional milestone for plotting other elements. To find the center of the hyperbola equation \[ \frac{(x+4)^2}{16} - \frac{(y-4)^2}{1} = 1 \], extract the values from the equation:
- The center \(h\) is derived from \(x + 4\), yielding \(h = -4\).
- The center \(k\) is derived from \(y - 4\), yielding \(k = 4\).
Vertices and Foci of Hyperbolas
Vertices and foci are crucial attributes of a hyperbola. The vertices are key points on each branch of the hyperbola and lie along the transverse axis.
The determination of the vertices in a hyperbola involves the distance \(a\), computed from the equation's denominator under the \(x\) term. For the equation \[ \frac{(x+4)^2}{16} - \frac{(y-4)^2}{1} = 1 \],
The determination of the vertices in a hyperbola involves the distance \(a\), computed from the equation's denominator under the \(x\) term. For the equation \[ \frac{(x+4)^2}{16} - \frac{(y-4)^2}{1} = 1 \],
- \(a^2 = 16\) implies \(a = 4\)
- The vertices are \((-4 \pm 4, 4)\) resulting in coordinates \((-8, 4)\) and \((0, 4)\)
- \(c^2 = 16 + 1 \) gives \( c = \sqrt{17} \)
- The foci points are stationed at \((-4 \pm \sqrt{17}, 4)\)
Asymptotes of Hyperbolas
Asymptotes are linear guides that indicate the direction of a hyperbola's branches. They are particularly significant because the two asymptotes form an "X-shape" where the hyperbola branches extend indefinitely yet never cross them.
For a horizontal hyperbola with an equation like \[ \frac{(x+4)^2}{16} - \frac{(y-4)^2}{1} = 1 \], the asymptote equations are derived from the formula:
For a horizontal hyperbola with an equation like \[ \frac{(x+4)^2}{16} - \frac{(y-4)^2}{1} = 1 \], the asymptote equations are derived from the formula:
- \(y = k \pm \frac{b}{a}(x - h)\)
- Plugging the given values, the equations become:
- \(y = 4 \pm \frac{1}{4}(x + 4)\)
