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Magazine Chapter 7: Problem 14
In Problems \(1-20\), find the center, foci, vertices, asymptotes, and eccentricity of the given hyperbola. Graph the hyperbola. $$ 10(x+1)^{2}-2\left(y-\frac{1}{2}\right)^{2}=100 $$
Short Answer
Expert verified
Center: \((-1, \frac{1}{2})\); Vertices: \((-1\pm\sqrt{10}, \frac{1}{2})\); Foci: \((-1\pm\sqrt{60}, \frac{1}{2})\); Eccentricity: \(\sqrt{6}\).
Step by step solution
01
Simplifying the Equation
The first step is to rewrite the hyperbola equation in the standard form. Start by dividing the given equation \[10(x+1)^{2}-2\left(y-\frac{1}{2}\right)^{2}=100\] by 100 to simplify it.The equation becomes:\[\frac{(x+1)^{2}}{10}-\frac{\left(y-\frac{1}{2}\right)^{2}}{50}=1\].This represents the standard form of a hyperbola with terms \(\frac{(x+1)^{2}}{a^2}\) and \(\frac{\left(y-\frac{1}{2}\right)^{2}}{b^2}\).
02
Identifying Center, Vertices and Foci Measurements
From the equation \[\frac{(x+1)^{2}}{10}-\frac{\left(y-\frac{1}{2}\right)^{2}}{50}=1\], we identify the center \((h, k) = (-1, \frac{1}{2})\).For the vertices, since the x-term is positive, it indicates a horizontal transverse axis: Vertices are at \(-1 \pm \sqrt{10}, \frac{1}{2}\).The foci are calculated using \(c^2 = a^2 + b^2\); hence, \(c^2 = 10 + 50 = 60\), which results in foci at \(-1 \pm \sqrt{60}, \frac{1}{2}\).
03
Calculating Asymptotes
The slopes of the asymptotes for a horizontal hyperbola are given by \(\pm \frac{b}{a}\).Here, \(a^2 = 10\) and \(b^2 = 50\) give us \( a = \sqrt{10}\) and \( b = \sqrt{50}\).The slopes of the asymptotes are \(\pm \frac{\sqrt{50}}{\sqrt{10}} = \pm \sqrt{5}\).The equations of the asymptotes are derived by using the point \((h, k) = (-1, \frac{1}{2})\) with the calculated slopes:\(y - \frac{1}{2} = \pm \sqrt{5}(x + 1)\).
04
Determining Eccentricity
The eccentricity \(e\) of a hyperbola is given by \(e = \frac{c}{a}\). We've previously computed \(c = \sqrt{60}\) and \(a = \sqrt{10}\), so\[e = \frac{\sqrt{60}}{\sqrt{10}} = \sqrt{6}\].
05
Graphing the Hyperbola
We have all required components to graph the hyperbola:- The center is at \((-1, \frac{1}{2})\).- The vertices are at \((-1+\sqrt{10}, \frac{1}{2})\) and \((-1-\sqrt{10}, \frac{1}{2})\).- The foci are at \((-1+\sqrt{60}, \frac{1}{2})\) and \((-1-\sqrt{60}, \frac{1}{2})\).- The asymptotes equations are \(y = \sqrt{5}(x + 1) + \frac{1}{2}\) and \(y = -\sqrt{5}(x + 1) + \frac{1}{2}\).Plot these points and draw the branches of the hyperbola which approach the asymptotes but never touch them.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Conic Sections
Conic sections are the curves obtained from the intersection of a plane with a double-napped cone. These fascinating curves include ellipses, circles, parabolas, and hyperbolas. Each conic section has unique properties depending on how the intersecting plane cuts through the cone. A hyperbola, in particular, is formed when the plane intersects both halves of the cone at an angle parallel to the cone's axis.
- Circle: A special type of ellipse where the plane cuts perpendicular to the cone's axis.
- Ellipse: Formed when the plane cuts through the cone at an angle, but not steep enough to reach both nappes.
- Parabola: Occurs when the plane is parallel to the slant of the cone, intersecting one nappe.
- Hyperbola: Arises when the plane intersects both nappes of the cone cyclically at a steep angle.
Eccentricity
Eccentricity is a measure that describes how much a conic section deviates from being circular. It is a crucial concept for understanding the shape and nature of conics. Eccentricity (denoted as \(e\)) is a non-negative parameter that gives insight into the curve's form.
- Circle: Eccentricity is \(0\) because a circle is perfectly round.
- Ellipse: Has eccentricity between \(0\) and \(1\). The closer \(e\) is to zero, the more circular it is.
- Parabola: Always has an eccentricity of \(1\).
- Hyperbola: Eccentricity is greater than \(1\). For our hyperbola, we calculated \(e = \sqrt{6}\), indicating it is more elongated as the conic deviates further from a circle.
Asymptotes
Asymptotes are straight lines that act as the "guiding" boundary lines for the branches of a hyperbola. These lines are crucial because while the hyperbola approaches them infinitely, it never intersects or crosses an asymptote. This property helps predict the hyperbola's behavior at infinity.For a hyperbola, asymptotes intersect at the hyperbola's center. The hyperbola appears to "open up" towards these lines. Here's a quick guide to calculating the asymptotes for a hyperbola:
- For a horizontal hyperbola centered at \((h, k)\), with equation \[(x-h)^2/a^2 - (y-k)^2/b^2 = 1\], the slopes of the asymptotes are \(\pm b/a\).
- These slopes determine the steepness and angle of the lines that the hyperbola approaches but never meets.
- In our hyperbola, \(a = \sqrt{10}\) and \(b = \sqrt{50}\) result in asymptotes with slopes of \(\pm \sqrt{5}\).
Foci
The foci (singular: focus) of a hyperbola are two distinct points located along the transverse axis, outside the vertices. They play an essential role in the geometric definition of a hyperbola. For any point on a hyperbola, the difference in distances to the two foci is constant and equal to the length of the transverse axis, \(2a\).In a hyperbola's equation \[(x-h)^2/a^2 - (y-k)^2/b^2 = 1\], the foci are calculated by finding \(c\) where \(c^2 = a^2 + b^2\). Here, \(c\) indicates how far the foci are from the center of the hyperbola:
- The foci for our solution are located at \((-1 + \sqrt{60}, \frac{1}{2})\) and \((-1 - \sqrt{60}, \frac{1}{2})\).
- These points are further from the hyperbola's center compared to the vertices, reflecting \(c > a\).
