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Magazine Chapter 10: Problem 66
Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections in Exercises \(57-68 .\) $$ x^{2}-y^{2}+4 x-6 y=6 $$
Short Answer
Expert verified
Center: (-2, -3), Vertices: (-3, -3) and (-1, -3), Foci: (-2卤鈭�2, -3), Asymptotes: y = x-1 and y = -x-5.
Step by step solution
01
Rewrite in Standard Form
To begin solving the problem, we must rewrite the given equation in a standard form of a hyperbola. Our equation is: \(x^{2} - y^{2} + 4x - 6y = 6\). We need to complete the square for both \(x\) and \(y\) terms.
02
Complete the Square
Rewrite the equation by rearranging and grouping: \((x^{2} + 4x) - (y^{2} + 6y) = 6\).Complete the square for \(x\):1. Take \(4/2 = 2\), square it to get \(4\).2. Add and subtract \(4\) inside the parentheses: \((x^{2} + 4x + 4 - 4)\).3. This becomes \((x + 2)^{2} - 4\).Complete the square for \(y\):1. Take \(-6/2 = -3\), square it to get \(9\).2. Add and subtract \(9\) inside the parentheses: \(-(y^{2} + 6y + 9 - 9)\).3. This becomes \(-(y + 3)^{2} + 9\).
03
Simplify the Equation
Substitute the completed squares back into the equation: \((x + 2)^2 - 4 - ((y + 3)^2 - 9) = 6\).Combine constants: \((x + 2)^2 - (y + 3)^2 = 6 + 4 - 9\).Simplify to get \((x + 2)^2 - (y + 3)^2 = 1\). This is the standard form of a hyperbola centered at \((-2, -3)\).
04
Identify the Center and Axes
The equation \((x + 2)^2 - (y + 3)^2 = 1\) represents a hyperbola centered at \((-2, -3)\) with its transverse axis along the \(x\)-direction.
05
Find Vertices and Foci
For a hyperbola \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\), the vertices are \((h \pm a, k)\).Here, \(h = -2\), \(k = -3\), \(a^2 = 1\) thus \(a = 1\).Vertices: \((-2 \pm 1, -3) = (-3, -3)\) and \((-1, -3)\).The foci are \((h \pm c, k)\) where \(c^2 = a^2 + b^2\). Since \(b = 1\), \(b^2 = 1\), \(c^2 = 2\), thus \(c = \sqrt{2}\).Foci: \((-2 \pm \sqrt{2}, -3)\).
06
Determine Asymptotes
The asymptotes of the hyperbola \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\) are given by the equations: \[ y - k = \pm \frac{b}{a}(x - h) \]Substituting the known values: \[ y + 3 = \pm 1 (x + 2) \]Thus, the asymptotes are \(y = x - 1\) and \(y = -x - 5\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Conic Sections
A conic section is a curve obtained by intersecting a cone with a plane. These shapes include circles, ellipses, parabolas, and hyperbolas. Each type has a unique equation and geometric properties.
Depending on the angle and position of the plane relative to the cone, the result is a different conic section.
Depending on the angle and position of the plane relative to the cone, the result is a different conic section.
- Circles and ellipses are formed when the plane cuts through only one nappe of the cone.
- Parabolas occur when the plane is parallel to a generating line of the cone.
- Hyperbolas are formed when the plane intersects both nappes of the cone.
Completing the Square
Completing the square is a method used to convert a quadratic equation into a form that reveals key characteristics like vertex or center of a conic section.In our exercise, we had the equation: \( x^2 - y^2 + 4x - 6y = 6 \).
To find its hyperbolic nature, we need to rewrite it in standard form by completing the square for both \( x \) and \( y \). This involves:
To find its hyperbolic nature, we need to rewrite it in standard form by completing the square for both \( x \) and \( y \). This involves:
- Reorganizing the terms: \((x^2 + 4x) - (y^2 + 6y) = 6\).
- Completing the square for \(x\): Add and subtract 4 within the equation: \((x + 2)^2 - 4\).
- Completing the square for \(y\): Add and subtract 9: \(-(y + 3)^2 + 9\).
Standard Form of a Hyperbola
The standard form of a hyperbola is crucial for understanding its features, such as its orientation, center, and asymptotes. For a hyperbola centered at \((h, k)\) oriented horizontally, the equation is:\[(x - h)^2/a^2 - (y - k)^2/b^2 = 1\]If oriented vertically, the equation is:\[(y - k)^2/a^2 - (x - h)^2/b^2 = 1\]In our exercise, after completing the square, the hyperbola's equation is:\[(x + 2)^2 - (y + 3)^2 = 1\]This indicates a horizontal orientation centered at \((-2, -3)\).
The values of \(a^2\) and \(b^2\) give insights into the hyperbola's shape. Here, both \(a^2\) and \(b^2\) are 1, indicating a symmetrically shaped hyperbola around its center.
The values of \(a^2\) and \(b^2\) give insights into the hyperbola's shape. Here, both \(a^2\) and \(b^2\) are 1, indicating a symmetrically shaped hyperbola around its center.
Vertices of a Hyperbola
Vertices are specific points on the hyperbola that lie on the axes and represent the 'widest' part of the curve. For a hyperbola in standard form, if it is horizontally aligned like ours, the vertices can be found at:
Knowing the vertices allows us to visualize the shape and orientation of the hyperbola more effectively.
- \((h \pm a, k)\)
- \((-3, -3)\)
- \((-1, -3)\)
Knowing the vertices allows us to visualize the shape and orientation of the hyperbola more effectively.
Asymptotes of a Hyperbola
Asymptotes are lines that a hyperbola approaches but never intersects. For hyperbolas, they aid in understanding the overall shape and direction of the branches.The equation for the asymptotes of a hyperbola in its standard form is:\[y - k = \pm \frac{b}{a}(x - h)\]For our equation \((x + 2)^2 - (y + 3)^2 = 1\), the asymptotes can be derived as follows:
- Substituting the values: \(y + 3 = \pm 1(x + 2)\)
- This simplifies to: \(y = x - 1\) and \(y = -x - 5\)
