Problem 17 - Precalculus: Mathematics for Calculus

Question

An equation of an ellipse is given. (a) Find the vertices, foci, and eccentricity of the ellipse. (b) Determine the lengths of the major and minor axes. (c) Sketch a graph of the ellipse. $$x^{2}+4 y^{2}=16$$

Step-by-Step Solution

Answer

Vertices: \((\pm 4, 0)\), Foci: \((\pm 2\sqrt{3}, 0)\), Eccentricity: \(\frac{\sqrt{3}}{2}\), Major axis: 8, Minor axis: 4.

Step 1: Standard Form

First, rewrite the given equation in the standard form of an ellipse. The general form of an ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). The given equation is \(x^2 + 4y^2 = 16\). To convert it to standard form, divide each term by 16 to get:\[\frac{x^2}{16} + \frac{y^2}{4} = 1\]This shows that \(a^2 = 16\) and \(b^2 = 4\), so \(a = 4\) and \(b = 2\).

Step 2: Find Vertices

The vertices of the ellipse are determined by the values of \(a\) and \(b\). For horizontal ellipses, the vertices are at \(\pm a\) along the x-axis. Thus, the vertices are at \((\pm 4, 0)\).

Step 3: Calculate Foci and Eccentricity

The foci are determined using the formula \(c^2 = a^2 - b^2\). Calculate \(c\):\[c^2 = 16 - 4 = 12 \Rightarrow c = \sqrt{12} = 2\sqrt{3}\]The foci are located at \((\pm c, 0)\), so they are \((\pm 2\sqrt{3}, 0)\). The eccentricity \(e\) is given by \(e = \frac{c}{a} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}\).

Step 4: Major and Minor Axes Lengths

The lengths of the major and minor axes are \(2a\) and \(2b\), respectively. Thus, the major axis is \(2 \times 4 = 8\) and the minor axis is \(2 \times 2 = 4\).

Step 5: Sketch the Graph

To sketch the ellipse, plot the vertices at \((4,0)\) and \((-4,0)\), the co-vertices at \((0,2)\) and \((0,-2)\), and the foci at \((2\sqrt{3}, 0)\) and \((-2\sqrt{3}, 0)\). Draw the ellipse centered at the origin, horizontally aligned, with the major axis along the x-axis.

Key Concepts

Vertices

The vertices of an ellipse are special points that help define its shape and orientation. Imagine these as the farthest points from the center along the longest diameter of the ellipse, known as the major axis.
In the standard form of the given ellipse equation, which is \(\frac{x^2}{16} + \frac{y^2}{4} = 1\), \(a\) and \(b\) represent the lengths of the semi-major and semi-minor axes, respectively. Here, \(a = 4\) and \(b = 2\).
Therefore, for a horizontal ellipse:

• The vertices are located along the x-axis at \((\pm a, 0)\).
• This gives us vertices at \((4, 0)\) and \((-4, 0)\).

These points are crucial for graphing the ellipse as they help define the overall dimensions.

Foci

The foci (singular: focus) of an ellipse are two fixed points located on the major axis, symmetrically about the center. They help define the curve such that the sum of the distances from any point on the ellipse to the two foci is constant.
Calculating the foci involves the formula \(c^2 = a^2 - b^2\), where \(c\) is the distance from the center to each focus. For our equation:

• \(c^2 = 16 - 4 = 12\)
• Thus, \(c = \sqrt{12} = 2\sqrt{3}\)

Given that this ellipse is horizontal, the foci are at \((\pm c, 0)\), specifically at \((2\sqrt{3}, 0)\) and \((-2\sqrt{3}, 0)\).
The position of the foci is key to understanding the ellipse's elongation.

Eccentricity

Eccentricity is a measure of how much an ellipse deviates from being circular. It is a dimensionless value, denoted by \(e\), and is calculated using the formula \(e = \frac{c}{a}\).
For a circle, the eccentricity is 0, as the foci converge at a single point (the center). For ellipses:

• The value of \(e\) ranges from 0 to 1.
• An ellipse closer to a circle has a lower eccentricity.
• A more elongated ellipse has a higher eccentricity.

In our case, substituting the values:

• \(e = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \approx 0.866\)

This high eccentricity indicates an ellipse that is noticeably stretched along the horizontal axis.
Understanding eccentricity helps distinguish different types of ellipses and interpret their physical or graphical implications.