Problem 7
Question
Match each conic section in Exercises \(5-8\) with one of these equations: $$\begin{array}{ll}{\frac{x^{2}}{4}+\frac{y^{2}}{9}=1,} & {\frac{x^{2}}{2}+y^{2}=1} \\ {\frac{y^{2}}{4}-x^{2}=1,} & {\frac{x^{2}}{4}-\frac{y^{2}}{9}=1}\end{array}$$ Then find the conic section's foci and vertices. If the conic section is a hyperbola, find its asymptotes as well. (GRAPH NOT COPY)
Step-by-Step Solution
Verified Answer
Match equations as follows: Ellipses: \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\), \(\frac{x^{2}}{2}+y^{2}=1\). Hyperbolas: \(\frac{y^{2}}{4}-x^{2}=1\), \(\frac{x^{2}}{4}-\frac{y^{2}}{9}=1\).
1Step 1: Identify Equation Type for Conic Sections
First, identify each equation as an ellipse, hyperbola, or parabola. An equation of the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) or \(\frac{y^2}{a^2} + \frac{x^2}{b^2} = 1\) is an ellipse. An equation of the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) or \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) is a hyperbola.
2Step 2: Match Conic Sections to Equations
Test the given equations against the standard forms. The equation \(\frac{x^2}{4} + \frac{y^2}{9} = 1\) is an ellipse (since both terms are positive). The equation \(\frac{x^2}{2} + y^2 = 1\) is also an ellipse. \(\frac{y^2}{4} - x^2 = 1\) and \(\frac{x^2}{4} - \frac{y^2}{9} = 1\) are hyperbolas (one term positive, one negative). Match these with the conic sections provided.
3Step 3: Find Vertices and Foci of Ellipses
For the ellipse \(\frac{x^2}{4} + \frac{y^2}{9} = 1\), calculate vertices as \((\pm 2, 0)\) and \((0, \pm 3)\). The foci are found using \(c = \sqrt{9-4} = \sqrt{5}\), giving foci at \((0, \pm \sqrt{5})\). For the ellipse \(\frac{x^2}{2} + y^2 = 1\), vertices are at \((\pm \sqrt{2}, 0)\) and \((0, \pm 1)\). Foci positions are \((0, \pm \sqrt{1}) = (0, \pm 1)\).
4Step 4: Find Vertices, Foci, and Asymptotes of Hyperbolas
For the hyperbola \(\frac{y^2}{4} - x^2 = 1\), vertices are \((0, \pm 2)\), foci are at \((0, \pm \sqrt{5})\). Asymptotes are \(y = \pm 2x\). For \(\frac{x^2}{4} - \frac{y^2}{9} = 1\), vertices are \((\pm 2, 0)\), foci are \((\pm \sqrt{13}, 0)\). Asymptotes are \(y = \pm \frac{3}{2}x\).
Key Concepts
EllipseHyperbolaFoci and Vertices
Ellipse
An ellipse is a special type of conic section that looks like an elongated circle or an oval shape. It's defined by the equation of the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). In this equation, the values of \(a\) and \(b\) determine the size and orientation of the ellipse. If \(a > b\), the ellipse is stretched along the x-axis, whereas if \(a < b\), it is elongated along the y-axis.
Ellipses have two sets of principal components:
Finding these points helps define the exact shape and position of the ellipse on a coordinate plane.
Ellipses have two sets of principal components:
- **Vertices:** These are the points at the farthest ends of the ellipse along its major axis. For an ellipse centered at the origin, vertices are at \((\pm a, 0)\) if the ellipse is horizontal, or \((0, \pm b)\) if vertical.
- **Foci:** Ellipses have two foci (singular: focus), which are located along the major axis, inside the curve. The distance to the foci can be found using \(c = \sqrt{a^2 - b^2}\).
Finding these points helps define the exact shape and position of the ellipse on a coordinate plane.
Hyperbola
A hyperbola is another type of conic section that appears as two separate curves, or branches. Much like reflecting in a mirror, hyperbolas are defined by equations of the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) or \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). The subtraction in their equations indicates the distinct, separate nature of their shape.
Key properties of a hyperbola include:
The properties of hyperbolas allow them to effectively model various physical phenomena, like the paths of certain celestial objects.
Key properties of a hyperbola include:
- **Vertices:** These are found at the closest points to the center of the hyperbola, located along the axis parallel to the first variable in the equation. For example, in \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), vertices are at \((\pm a, 0)\).
- **Foci:** These are special points used to define the bifurcating nature of the hyperbola, found using \(c = \sqrt{a^2 + b^2}\).
- **Asymptotes:** These are lines that the branches of the hyperbola approach but never touch, described by \(y = \pm \frac{b}{a}x\) in the horizontal form, or \(x = \pm \frac{a}{b}y\) if vertical.
The properties of hyperbolas allow them to effectively model various physical phenomena, like the paths of certain celestial objects.
Foci and Vertices
The terms **foci** and **vertices** are fundamental in understanding the geometry of both ellipses and hyperbolas.
**Vertices** refer to the significant points on the graph of a conic section. In ellipses, they're the extremal points on the long or major axis, while in hyperbolas, vertices indicate the turning points of the branches.
For **foci**, these are particular points inside the conic sections that define their unique properties. By fixing these points and varying the distance, one can draw an ellipse or hyperbola.
Understanding these elements is crucial to mastering the characteristics and equations of conic sections.
**Vertices** refer to the significant points on the graph of a conic section. In ellipses, they're the extremal points on the long or major axis, while in hyperbolas, vertices indicate the turning points of the branches.
For **foci**, these are particular points inside the conic sections that define their unique properties. By fixing these points and varying the distance, one can draw an ellipse or hyperbola.
- In ellipses: Distance from any point on the ellipse to both foci remains constant. It equals the length of the major axis, hence why ellipses can be described by their constrant sumproperty.
- In hyperbolas: The focus helps define the curve's shape and direction. The difference in distances from any point on the hyperbola to its foci is constant.
Understanding these elements is crucial to mastering the characteristics and equations of conic sections.
Other exercises in this chapter
Problem 7
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