Problem 25
Question
Exercises 25 and 26 give information about the foci and vertices of ellipses centered at the origin of the \(x y\) -plane. In each case, find the ellipse's standard-form equation from the given information. $$ \begin{array}{l}{\text { Foci: }( \pm \sqrt{2}, 0)} \\ {\text { Vertices: }( \pm 2,0)}\end{array} $$
Step-by-Step Solution
Verified Answer
\(\frac{x^2}{4} + \frac{y^2}{2} = 1\)
1Step 1: Understanding the Paper Setup
For the given problem, we know the foci and vertices of an ellipse centered at the origin of the xy-plane. We are tasked with finding the standard form equation of this ellipse. Given the foci are \((\pm \sqrt{2}, 0)\) and the vertices are \((\pm 2, 0)\), we can ascertain that the ellipse is horizontally oriented because both the foci and vertices are along the x-axis.
2Step 2: Write the Standard Form for a Horizontal Ellipse
The standard form of a horizontally oriented ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a\) is the semi-major axis, \(b\) is the semi-minor axis, and the distance from the center to each focus \(c\) satisfies the relationship \(c^2 = a^2 - b^2\).
3Step 3: Identify the Values of a and c
From the vertices \((\pm 2, 0)\), we get \(a = 2\). From the foci \((\pm \sqrt{2}, 0)\), we have \(c = \sqrt{2}\).
4Step 4: Calculate b using the Relationship with a and c
Using the relationship \(c^2 = a^2 - b^2\):\[(\sqrt{2})^2 = 2^2 - b^2\]\[2 = 4 - b^2\]\[b^2 = 2\]
5Step 5: Write Down the Equation of the Ellipse
Substituting the values of \(a^2\) and \(b^2\) into the standard form of the ellipse, we get:\[\frac{x^2}{4} + \frac{y^2}{2} = 1\]
Key Concepts
Foci of an EllipseStandard Form of an EllipseEllipse Centered at the OriginVertices of an Ellipse
Foci of an Ellipse
The foci of an ellipse are two specific points located along the major axis of the ellipse. They have a unique property that the sum of the distances from any point on the ellipse to the two foci is constant. This characteristic defines the shape of the ellipse.
In our given problem, the foci are located at \((\pm \sqrt{2}, 0)\). This tells us that the ellipse is centered around the origin and is horizontally oriented since the foci lie on the x-axis. The distance from the center to each focus, \(c\), is \(\sqrt{2}\). Understanding where the foci are positioned helps in determining the ellipse's configuration and aids in finding the standard form of the ellipse equation.
In our given problem, the foci are located at \((\pm \sqrt{2}, 0)\). This tells us that the ellipse is centered around the origin and is horizontally oriented since the foci lie on the x-axis. The distance from the center to each focus, \(c\), is \(\sqrt{2}\). Understanding where the foci are positioned helps in determining the ellipse's configuration and aids in finding the standard form of the ellipse equation.
Standard Form of an Ellipse
The standard form of an ellipse equation depends on the orientation of the ellipse. If the ellipse is horizontal, the standard form is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Here, \(a\) represents the semi-major axis (the longest radius), and \(b\) represents the semi-minor axis (the shortest radius).
In our exercise, given the ellipse is horizontal, we use this equation to represent the ellipse. This equation is instrumental in encapsulating the properties of the ellipse, linking the axes' lengths. Additionally, it helps us understand the symmetrical nature of the ellipse about its center. It forms the basis of analyzing any ellipse mathematically.
In our exercise, given the ellipse is horizontal, we use this equation to represent the ellipse. This equation is instrumental in encapsulating the properties of the ellipse, linking the axes' lengths. Additionally, it helps us understand the symmetrical nature of the ellipse about its center. It forms the basis of analyzing any ellipse mathematically.
Ellipse Centered at the Origin
An ellipse centered at the origin means the midpoint between its foci is at point \((0,0)\). In our scenario, the foci’s positions and the vertices confirm that the ellipse is centered at the origin.
Placing the ellipse at the origin simplifies its equation, as the center coinciding with the origin removes the need for additional terms in the equation that would account for shifts along the x or y axes. Simply put, it lets us use the basic form of elliptical equations without any modification for relocation.
Placing the ellipse at the origin simplifies its equation, as the center coinciding with the origin removes the need for additional terms in the equation that would account for shifts along the x or y axes. Simply put, it lets us use the basic form of elliptical equations without any modification for relocation.
Vertices of an Ellipse
The vertices of an ellipse are points lying at the ends of the major axis. In essence, they represent the farthest points from the center on the ellipse. For our exercise, the vertices are located at \((\pm 2, 0)\).
This tells us several critical things: firstly, that the semi-major axis length \(a\) is 2, since it represents half the distance between the vertices, and secondly, this confirms the ellipse's horizontal orientation.
This tells us several critical things: firstly, that the semi-major axis length \(a\) is 2, since it represents half the distance between the vertices, and secondly, this confirms the ellipse's horizontal orientation.
- The vertices help in determining the ellipse's size.
- From the vertices, we can compute part of the necessary parameters to form the ellipse's standard equation.
Other exercises in this chapter
Problem 25
Find polar equations for the circles in Exercises \(21-28 .\) Sketch each circle in the coordinate plane and label it with both its Cartesian and polar equation
View solution Problem 25
Sketch the region defined by the inequalities \(-1 \leq r \leq 2\) and \(-\pi / 2 \leq \theta \leq \pi / 2 .\)
View solution Problem 25
Replace the polar equations in Exercises \(23-48\) by equivalent Cartesian equations. Then describe or identify the graph. $$ r \sin \theta=0 $$
View solution Problem 25
In Exercises \(23-30\) , find the eccentricity of the hyperbola. Then find and graph the hyperbola's foci and directrices. $$ y^{2}-x^{2}=8 $$
View solution