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Find the vertices, foci, and eccentricity of the ellipse. Determine the lengths of the major and minor axes, and sketch the graph. $$x^{2}+4 y^{2}=1$$

Short Answer

Expert verified
Vertices: (1, 0), (-1, 0); Foci: (±√3/2, 0); Eccentricity: √3/2. Major axis: 2; Minor axis: 1.

Step by step solution

01

Identify the Standard Form of an Ellipse

The given equation is \( x^2 + 4y^2 = 1 \). To identify it as an ellipse, rewrite it in standard form. Divide every term by 1 to match the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). In this case, the equation is already in a suitable form: \[ \frac{x^2}{1} + \frac{y^2}{(1/2)^2} = 1 \] Here, \( a = 1 \) and \( b = \frac{1}{2} \). This is an ellipse centered at the origin with the x-axis as the major axis.
02

Calculate the Vertices

The vertices of the ellipse are located along the major axis. Since the major axis is aligned with the x-axis (as \( a > b \)), the vertices \( V \) can be found at \( (\pm a, 0) \). Here, \( a = 1 \), so the vertices are:\[ V_1 = (1, 0) \quad \text{and} \quad V_2 = (-1, 0) \]
03

Determine the Foci

The foci of the ellipse are also located along the major axis. The distance from the center to each focus is given by \( c = \sqrt{a^2 - b^2} \). First, calculate \( c \):\[ c = \sqrt{1^2 - \left(\frac{1}{2}\right)^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \] Thus, the foci \( F \) are located at: \[ F_1 = \left( \frac{\sqrt{3}}{2}, 0 \right) \quad \text{and} \quad F_2 = \left( -\frac{\sqrt{3}}{2}, 0 \right) \]
04

Calculate the Eccentricity

The eccentricity \( e \) of the ellipse is calculated using the formula \( e = \frac{c}{a} \). Substituting the known values:\[ e = \frac{\frac{\sqrt{3}}{2}}{1} = \frac{\sqrt{3}}{2} \] The eccentricity \( e \) is therefore \( \frac{\sqrt{3}}{2} \), indicating the shape of the ellipse is non-circular.
05

Determine the Lengths of the Major and Minor Axes

The lengths of the major and minor axes are calculated as follows:- The length of the major axis is \( 2a \), which is \( 2 \times 1 = 2 \).- The length of the minor axis is \( 2b \), which is \( 2 \times \frac{1}{2} = 1 \).Thus, the major axis measures 2 units, and the minor axis measures 1 unit.
06

Sketch the Graph

For sketching:- Plot the vertices \((1,0)\) and \((-1,0)\) on the x-axis, as they define the extent of the major axis.- Plot points at \( (0, 0.5) \) and \( (0, -0.5) \) on the y-axis, which represent the extent of the minor axis.- Connect these points smoothly to form an elongated oval shape centered at the origin.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertices
The vertices of an ellipse are essential as they highlight the extent of the ellipse along its major axis. **Vertices** are points where the ellipse intersects its major axis, marking the outermost limits on that line. You find the vertices by looking at the term with the larger denominator in the standard form equation of an ellipse, \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\,\, (a > b).\] In this exercise, we have identified the major axis along the x-axis due to \(a > b\). The vertices lie at \((\pm a, 0)\). Given \(a = 1\), the vertices are \((1, 0)\) and \((-1, 0)\). These points define how far the ellipse stretches along the x-direction.
Foci
The foci of an ellipse play a crucial role in its geometric definition and its structure. An ellipse is defined as the set of points such that the sum of the distances to two fixed points, known as **foci**, is constant. These foci are always positioned on the major axis, making them pivotal for the shape and nature of the ellipse. The distance from the center to each focus is represented by \(c\), and it can be calculated using the formula:\[c = \sqrt{a^2 - b^2}.\]In our exercise, after calculations, we find \(c = \frac{\sqrt{3}}{2}\). Therefore, the foci are located at \(\left( \pm \frac{\sqrt{3}}{2}, 0 \right)\). These points contribute to the unique property that defines the curve.
Eccentricity
The concept of eccentricity in an ellipse conveys its "flatness" or deviation from being a perfect circle. Eccentricity, denoted by \(e\), is calculated using the relationship:\[e = \frac{c}{a},\]where \(c\) is the distance from the center to either focus. The value of \(e\) is always between 0 and 1 for ellipses; a lower value signifies a shape closer to a circle, while a higher value suggests a more elongated form. For the ellipse in this exercise, the eccentricity is found to be \(\frac{\sqrt{3}}{2}\). This indicates a distinctly oval shape, confirming the ellipse isn't a circle since the eccentricity is not zero.
Major Axis
Understanding the major axis is vital as it is the longest diameter that passes through the center of the ellipse. It lies along the direction that maximizes the spread of the ellipse, housing both vertices and foci. For this particular ellipse, the major axis is identified along the x-axis because \(a > b\). The length is calculated as:\[2a.\]Given \(a = 1\), the major axis measures \(2\) units in total. This axis not only centralizes the geometric symmetry of the ellipse but also delineates the path along which the maximum distance is achieved, from one vertex, through the center, to the opposite vertex.
Minor Axis
The minor axis of an ellipse characterizes the minimum width of the shape, cutting across the center perpendicular to the major axis. Although shorter, it's equally essential to describe the ellipse's overall breadth. The length of the minor axis can be determined using the formula:\[2b.\]In our case, \(b = \frac{1}{2}\), rendering the total length of the minor axis as \(1\) unit. This axis highlights the compactness of the ellipse, contrasting with the extension defined by the major axis. Points along the minor axis ensure the ellipse remains constrained, maintaining a balance that deviates from circularity but holds to ellipse properties.

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