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Magazine Chapter 10: Problem 14
Sketch the hyperbola, and label the vertices, foci, and asymptotes. (a) \(x^{2}-4 y^{2}+2 x+8 y-7=0\) (b) \(16 x^{2}-y^{2}-32 x-6 y=57\)
Short Answer
Expert verified
Centers: Part A at (-1,1), Part B at (1,-3). Vertices, foci, and asymptotes identified and sketched.
Step by step solution
01
Rewrite in Standard Form (Part A)
For the equation \(x^2 - 4y^2 + 2x + 8y - 7 = 0\), start by grouping \(x\) and \(y\) terms separately: \((x^2 + 2x) - (4y^2 - 8y) = 7\). Complete the square to simplify this into its standard hyperbola form. \(x^2 + 2x\):\(x^2 + 2x = (x+1)^2 - 1\).For \(-4(y^2 - 2y)\), factor out \(-4\) then complete the square: \(y^2 - 2y = (y-1)^2 - 1\), so, \(-4((y-1)^2 - 1) = -4(y-1)^2 + 4\).Insert back, rearranging gives \((x+1)^2 - 4(y-1)^2 = 12\). Divide all terms by 12 to normalize:\[\frac{(x+1)^2}{12} - \frac{(y-1)^2}{3} = 1\].
02
Identify Key Features (Part A)
In the standard form \( \frac{(x+1)^2}{12} - \frac{(y-1)^2}{3} = 1\), we can identify that this is a horizontal hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) centered at \((-1,1)\) with \(a^2 = 12\) and \(b^2 = 3\).Vertices are at \( (-1\pm \sqrt{12}, 1)\) or roughly \((-1\pm 3.4641, 1)\): \((-4.4641, 1)\) and \((2.4641, 1)\).Foci are calculated using \( c^2 = a^2 + b^2 = 15\), thus \(c = \sqrt{15} \approx 3.8729\).Foci: \((-1\pm 3.8729, 1)\): \((-4.8729, 1)\) and \((2.8729, 1)\).Asymptotes have the equations: \(y - 1 = \pm \frac{b}{a}(x + 1)\) i.e., \(y - 1 = \pm \frac{1}{2\sqrt{3}}(x + 1)\).
03
Rewrite in Standard Form (Part B)
For the equation \(16x^2 - y^2 - 32x - 6y = 57\), start with grouping \(x\) and \(y\) terms: \(16(x^2 - 2x) - (y^2 + 6y) = 57\).Complete the square for both:\(x^2 - 2x = (x-1)^2 - 1\), and then: \(16((x-1)^2 - 1) = 16(x-1)^2 - 16\).\(y^2 + 6y = (y+3)^2 - 9\), leads to \(-(y+3)^2 + 9\).Simply this into standard form: \(16(x-1)^2 - (y+3)^2 = 64\).Divide by 64: \[\frac{(x-1)^2}{4} - \frac{(y+3)^2}{64} = 1\].
04
Identify Key Features (Part B)
The standard form \(\frac{(x-1)^2}{4} - \frac{(y+3)^2}{64} = 1\) represents a horizontal hyperbola centered at \((1, -3)\) with \(a^2 = 4\), \(b^2 = 64\).Vertices are at \( (1\pm a, -3)\) or \( (1\pm 2, -3)\): \((-1, -3)\) and \((3, -3)\).Calculate foci using \(c^2 = a^2 + b^2 = 68\), therefore \(c = \sqrt{68} \approx 8.2462\).Foci: \((1\pm 8.2462, -3)\): \((-7.2462, -3)\) and \((9.2462, -3)\).Asymptotes have equations: \(y + 3 = \pm \frac{b}{a}(x - 1)\) which simplifies to: \(y + 3 = \pm 4(x - 1)\).
05
Sketch Hyperbolas and Label
For both hyperbolas, draw: For Part A: Center \((-1, 1)\), vertices at \((-4.46, 1)\) & \((2.46, 1)\), foci at \((-4.87, 1)\) & \((2.87, 1)\), and asymptotes \(y - 1 = \pm \frac{1}{2\sqrt{3}}(x + 1)\).For Part B: Center \((1, -3)\), vertices at \((-1, -3)\) & \((3, -3)\), foci at \((-7.25, -3)\) & \((9.25, -3)\), and asymptotes \(y + 3 = \pm 4(x - 1)\). Indicate each vertex, focus, and draw the asymptotes clearly to explore the shape.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertices of Hyperbola
When graphing a hyperbola, identifying its vertices is crucial as they help in determining the overall shape and position of the graph. The vertices are the points where the hyperbola intersects its principal axis. To find them, you need the hyperbola's standard form.
For example, consider a hyperbola of the form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). This is a horizontal hyperbola. Here, \((h, k)\) represents the center, while \(a^2\) and \(b^2\) are constants that aid in calculating the ellipse dimensions.
For example, consider a hyperbola of the form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). This is a horizontal hyperbola. Here, \((h, k)\) represents the center, while \(a^2\) and \(b^2\) are constants that aid in calculating the ellipse dimensions.
- For horizontal hyperbolas, the vertices are located at \((h\pm a, k)\).
- For vertical hyperbolas, the vertices switch to \((h, k\pm a)\).
Foci of Hyperbola
The foci of a hyperbola are two key points located symmetrically along its principal axis, similar to the ellipse's foci. They are crucial for defining the hyperbola's shape and provide an anchor for its characteristic property鈥攅very point on the hyperbola has a constant difference of distances to the two foci.
To find the foci, use the relationship \(c^2 = a^2 + b^2\) where \(c\) indicates the distance from the center to each focus.
To find the foci, use the relationship \(c^2 = a^2 + b^2\) where \(c\) indicates the distance from the center to each focus.
- For a horizontal hyperbola \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the foci are positioned at \((h\pm c, k)\).
- For a vertical hyperbola \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), the foci are at \((h, k\pm c)\).
Asymptotes of Hyperbola
Asymptotes are lines that the hyperbola approaches but never quite reaches. They give you a skeletal framework of your hyperbola, providing the orientation and the 'opening' direction of its curves. For hyperbolas, asymptotes run through the center and are crucial for accurate graphing.
Here's how you calculate them:
Here's how you calculate them:
- For horizontal hyperbolas \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the asymptotes are expressed as \(y-k = \pm \frac{b}{a}(x-h)\).
- For vertical hyperbolas, \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), the equations become \(y-k = \pm \frac{a}{b}(x-h)\).
