Problem 36 - Algebra and Trigonometry Real Mathematics, Real People

Question

(a) find the standard form of the equation of the ellipse, (b) find the center, vertices, foci, and eccentricity of the ellipse, and (c) sketch the ellipse. Use a graphing utility to verify your graph. $$16 x^{2}+y^{2}=16$$

Step-by-Step Solution

Answer

The standard form of the ellipse is \(x^{2} + \frac{y^{2}}{16} = 1\). The center is at (0, 0), the vertices are at (±1, 0), and there are no foci because the ellipse is a circle with radius 1.

Step 1: Convert the Equation into Standard Form

The given equation of the ellipse is \(16x^{2} + y^{2} = 16\). The standard form of an ellipse with the major axis along the x-axis is \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\). We should divide every term of the given equation by 16 to obtain the standard form which results in \(\frac{x^{2}}{1} + \frac{y^{2}}{16} = 1 \). Hence, \(a^{2} = 1\) and \(b^{2} = 16\).

Step 2: Find the Center, Vertices, Foci, and Eccentricity of the Ellipse

The center of the ellipse is given by the origin (0, 0) because the terms with x and y in the equation has no added or subtracted constant. The vertices are at (±a, 0) = (±1, 0). The co-vertices lie along the y-axis at (0, ±b) = (0, ±4). The foci are at (±c, 0) where \(c = \sqrt{a^{2} - b^{2}} = \sqrt{1 - 16}\) which is not a real number. In this case, the ellipse is a circle with radius 1 and no foci. The eccentricity which is defined by \(\frac{c}{a}\) does not exist because c does not exist.

Step 3: Illustrate the Ellipse

Draw an x-y cartesian coordinate system. Mark the center at the origin. Plot the vertices at (±1, 0) and the co-vertices at (0, ±4). The drawn ellipse is in fact a circle because a=b and eccentricity=0. Make sure the circle is centered at the origin and touches the vertices and co-vertices.

Key Concepts

Standard Form of Ellipse

Ellipses have a characteristic equation that can be transformed into what we call "standard form." This form helps us effortlessly identify properties such as the center, vertices, and axes lengths. For an ellipse with a horizontal major axis, the standard form is:

• \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\)

Where \(a\) denotes the semi-major axis length along the x-axis and \(b\) represents the semi-minor axis length along the y-axis.
To convert a given ellipse equation into its standard form, each term needs to be divided by the total constant. In our case, from the equation \(16x^{2} + y^{2} = 16\), we've divided all terms by 16, resulting in \(\frac{x^{2}}{1} + \frac{y^{2}}{16} = 1\). Now, we can easily identify \(a^{2} = 1\) and \(b^{2} = 16\). This method unveils the essential parameters of the ellipse.

Center of an Ellipse

The center of an ellipse is the point around which the ellipse is symmetrically placed. In its standard form, if there are no added or subtracted terms involving x or y, the ellipse's center is at the origin.
The equation \(16x^{2} + y^{2} = 16\) converts to \(\frac{x^{2}}{1} + \frac{y^{2}}{16} = 1\) after standardizing, assuring that the center is \((0, 0)\) as there are no shifted constants.
Identifying the center is crucial as it allows us to properly position the ellipse on a graph.

Vertices of an Ellipse

Vertices of an ellipse are the points where the ellipse intersects the major and minor axes. These points help define the shape and size of the ellipse.

• If the major axis is horizontal, as in this case, then the vertices are located at \((\pm a, 0)\).
• For the equation \(\frac{x^{2}}{1} + \frac{y^{2}}{16} = 1\), the vertices are \((\pm 1, 0)\) as \(a = \sqrt{1} = 1\).
• Co-vertices along the y-axis are at \((0, \pm 4)\) calculated from \(b = \sqrt{16} = 4\).

This positioning clearly separates the primary axis (horizontal) from the secondary axis (vertical), setting the structure for our graph.

Eccentricity

Eccentricity determines how much an ellipse deviates from being a perfect circle. It is defined as the ratio \(\frac{c}{a}\), where \(c\) is the distance from the center to each focus. For a traditional ellipse, eccentricity ranges between 0 and 1.
However, in this exercise, as we try to calculate \(c\) through the formula \(c = \sqrt{b^{2} - a^{2}}\), we find \(\sqrt{16 - 1} = \sqrt{15}\). This results in an eccentricity of \(\frac{\sqrt{15}}{1}\), giving an undefined reality to this eccentricity as it surpasses traditional bounds. This suggests our ellipse behaves more like a circle with a notably small eccentricity close to zero.
Understanding eccentricity provides insight into the curvature and roundness of the ellipse.

Graphing Ellipses

Drawing an ellipse accurately requires understanding its key components: the center, vertices, axes, and eccentricity.

• Begin by marking the center on the coordinate system—here, at \((0, 0)\).
• Position the vertices along the x-axis at \((\pm 1, 0)\), and the co-vertices along the y-axis at \((0, \pm 4)\).
• The locus of these points forms the boundary of our ellipse, which resembles a circle, given \(a = b\).

In this configuration, despite minimal eccentricity, it has equal semi-major and semi-minor axes. Remember, knowing how to plot these key points ensures an accurate representation of an ellipse.