Problem 9 - Precalculus: Mathematical for Calculus

Question

Find the vertices, foci, and eccentricity of the ellipse. Determine the lengths of the major and minor axes, and sketch the graph. $$x^{2}+4 y^{2}=16$$

Step-by-Step Solution

Answer

Vertices: (-4,0) and (4,0); Foci: (-2√3,0) and (2√3,0); Eccentricity: √3/2; Major axis: 8; Minor axis: 4.

Step 1: Identify the Standard Form

The given equation is \(x^2 + 4y^2 = 16\). This resembles an ellipse in the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) but it needs to be transformed to the standard form first. Divide the entire equation by 16 to get: \[\frac{x^2}{16} + \frac{y^2}{4} = 1.\] This confirms it's an ellipse with \(a^2 = 16\) and \(b^2 = 4\).

Step 2: Determine the Lengths of Axes

In the standard form of an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), \(a\) is the length of the semi-major axis and \(b\) is the length of the semi-minor axis. Here, \(a^2 = 16\) which implies \(a = 4\), while \(b^2 = 4\) implies \(b = 2\). Therefore, the lengths of the major and minor axes are \(2a = 8\) and \(2b = 4\), respectively.

Step 3: Identify the Orientation

Since \(a > b\), the ellipse is oriented along the x-axis, meaning the major axis is horizontal and centered at the origin (0,0).

Step 4: Find the Vertices

The vertices are located at \((-a, 0)\), \((a, 0)\) due to the horizontal orientation. Substituting \(a = 4\), the vertices are at \((-4, 0)\) and \((4, 0)\).

Step 5: Calculate the Foci

The foci \((c)\) are given by the formula \(c^2 = a^2 - b^2\). Substitute \(a^2 = 16\) and \(b^2 = 4\) to find \(c^2 = 12\), so \(c = \sqrt{12}\) or \(c = 2\sqrt{3}\). The foci are at \((-2\sqrt{3}, 0)\) and \((2\sqrt{3}, 0)\).

Step 6: Determine Eccentricity

Eccentricity \(e\) of an ellipse is given by \(e = \frac{c}{a}\). Substituting \(c = 2\sqrt{3}\) and \(a = 4\), the eccentricity is \(e = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}\).

Step 7: Sketch the Ellipse

On a coordinate plane, draw a horizontal ellipse centered at the origin. The major axis is 8 units along the x-axis between (-4, 0) and (4, 0). The minor axis is 4 units along the y-axis between (0, -2) and (0, 2). Highlight the foci at approximately (-3.46, 0) and (3.46, 0).

Key Concepts

Vertices

In an ellipse, vertices are the points where the ellipse is widest and those points lie along the major axis. For the ellipse described by \(\frac{x^2}{16} + \frac{y^2}{4} = 1\), the vertices are determined by the value of \(a\), which is the semi-major axis length. Here, \(a = 4\), and since the ellipse is oriented horizontally, the vertices are at the coordinates \((-4, 0)\) and \(4, 0)\).Remember:

• The vertices lie on the major axis.
• They are equidistant from the center, in this case at the origin \(0, 0\).

Finding vertices is a key step in sketching and understanding the shape and orientation of the ellipse.

Foci

The foci of an ellipse are two distinct points located along the major axis. They are closer to the center compared to the vertices and are crucial in defining the curve of an ellipse. The distance from the center to each focus can be calculated with the formula \(c^2 = a^2 - b^2\), where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively.In our case:

• \(a = 4\) and \(b = 2\), so \((a^2 = 16, b^2 = 4)\).
• Then \(c^2 = 16 - 4 = 12\), resulting in \(c = \sqrt{12} = 2\sqrt{3}\).

The foci coordinates are therefore \((-2\sqrt{3}, 0)\) and \(2\sqrt{3}, 0)\). Placing the foci correctly is essential to drawing an accurate ellipse, as any point on the ellipse has the same total distance to the foci.

Eccentricity

Eccentricity is a measure that describes how "stretched" an ellipse is. It is a dimensionless number ranging from 0 to 1 for an ellipse, where 0 represents a perfectly circular shape and values closer to 1 indicate a more elongated shape. The formula to find eccentricity \(e\) is \(e = \frac{c}{a}\), where \(c\) is the distance from the center to a focus and \(a\) is the length of the semi-major axis.In this ellipse:

• \(c = 2\sqrt{3}\) and \(a = 4\).
• The eccentricity is: \(e = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}\).

Understanding the eccentricity helps in grasping the nature of the ellipse's geometry, being a key attribute of its shape.

Major Axis

The major axis is the longer diameter of an ellipse, stretching across the widest part of the shape. In the standard form of an ellipse equation, it is defined by the term with the larger denominator—in this case, \(\frac{x^2}{16}\), meaning the major axis is along the x-axis.For this ellipse:

• The major axis length is \(2a = 8\) because \(a = 4\).

The endpoints of the major axis are the vertices \((-4, 0)\) and \((4, 0)\), illustrating the full span of the ellipse in that direction. This axis is crucial for accurately drawing and understanding the ellipse's proportions.

Minor Axis

The minor axis of an ellipse is its shortest diameter, perpendicular to the major axis. It stretches across the narrower part of the ellipse. For the given ellipse, the minor axis lies along the y-axis because the denominator of the \(y^2\) term is smaller.Calculations for the minor axis include:

• \(b = 2\), which gives a total minor axis length of \(2b = 4\).
• This means it stretches between points (0, -2) and (0, 2) as it is centered at the origin.

The minor axis aids in outlining the shape of the ellipse and, together with the major axis, defines its overall proportions and size.