Problem 19 - 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. $$16 x^{2}+25 y^{2}=1600$$

Step-by-Step Solution

Answer

Vertices: \((-10, 0)\) and \((10, 0)\), foci: \((-6, 0)\) and \((6, 0)\), eccentricity: 0.6. Major axis: 20, minor axis: 16.

Step 1: Convert the Equation to Standard Form

The given equation is \( 16x^{2} + 25y^{2} = 1600 \). Divide the entire equation by 1600 to convert it to standard form:\ \[ \frac{x^{2}}{100} + \frac{y^{2}}{64} = 1 \]\ This can be rewritten as \[ \frac{x^{2}}{10^{2}} + \frac{y^{2}}{8^{2}} = 1 \], indicating a horizontal major axis since \(a^2 > b^2\).

Step 2: Identify Semi-Major and Semi-Minor Axes

The standard form of the ellipse is \( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \) with \(a^{2} = 100\) and \(b^{2} = 64\). Therefore, \(a = 10\) and \(b = 8\). The larger value, \(a\), indicates the semi-major axis, and \(b\) is the semi-minor axis.

Step 3: Find the Vertices

The ellipse has its center at the origin \((0, 0)\). Since the major axis is horizontal, the vertices are at \((-a, 0)\) and \((a, 0)\). With \(a = 10\), the vertices are \((-10, 0)\) and \((10, 0)\).

Step 4: Calculate the Foci

The distance of each focus from the center is given by \(c = \sqrt{a^{2} - b^{2}}\). Calculate \(c\):\\[ c = \sqrt{10^{2} - 8^{2}} = \sqrt{36} = 6 \] \Thus, the foci are \((-6, 0)\) and \((6, 0)\).

Step 5: Determine the Eccentricity

The eccentricity of an ellipse is given by \(e = \frac{c}{a}\). Substitute the values:\\[ e = \frac{6}{10} = 0.6 \].

Step 6: Calculate the Lengths of Major and Minor Axes

The length of the major axis is \(2a\), which is \(2 \times 10 = 20\). The length of the minor axis is \(2b\), which is \(2 \times 8 = 16\).

Step 7: Sketch the Ellipse

To sketch the ellipse, plot the vertices \((-10, 0)\), \((10, 0)\), and minor axis endpoints \((0, -8)\), \((0, 8)\). Draw a smooth, elongated ellipse through these points centered at the origin, with the horizontal span longer than the vertical one.

Key Concepts

Vertices of Ellipse

In an ellipse, the vertices are the points where the ellipse is longest along its major axis. Think of these as the most distanced points from each other across the ellipse. For the ellipse given in the equation \(\frac{x^{2}}{100} + \frac{y^{2}}{64} = 1\), we have learned that the major axis is horizontal because the coefficient of \(x^2\) is larger. The formula for finding vertices in an ellipse centered at the origin is

• Horizontal ellipse: \((-a, 0)\) and \((a, 0)\)
• Vertical ellipse: \((0, -a)\) and \((0, a)\)

In this case, since \(a = 10\), the vertices are precisely at the points \((-10, 0)\) and \((10, 0)\). This means the ellipse stretches 10 units in both directions from the center, making the distance between the vertices 20 units in total.

Foci of Ellipse

The foci (pronounced "fo-sigh") are critical points inside the ellipse. They have a unique property: the sum of the distances from any point on the ellipse to the two foci is constant. For the ellipse described as \(\frac{x^{2}}{10^{2}} + \frac{y^{2}}{8^{2}} = 1\), the foci are located along the major axis, inside the ellipse. To find them, use the formula:

• \(c = \sqrt{a^2 - b^2}\)

For this ellipse, \(a^2 = 100\) and \(b^2 = 64\), giving \(c = \sqrt{100 - 64} = \sqrt{36} = 6\). Therefore, since the major axis is horizontal:

• The foci are at \((-6, 0)\) and \((6, 0)\)

These points lie within the ellipse, 6 units away from the center along the x-axis.

Eccentricity of Ellipse

Eccentricity is a measure that describes how elongated an ellipse is compared to a perfect circle. It's represented by \(e\) and defined for an ellipse as: \[ e = \frac{c}{a} \]where \(c\) is the distance from the center to a focus, and \(a\) is the distance from the center to a vertex (semi-major axis). For the given ellipse, \(c = 6\) and \(a = 10\), so the eccentricity is: \[ e = \frac{6}{10} = 0.6 \]An eccentricity of 0 characterizes a perfect circle, while one closer to 1 indicates a more elongated shape. Therefore, an eccentricity of 0.6 means this ellipse is moderately elongated but not overly so. This value tells you how much the ellipse deviates from being circular.

Major and Minor Axes

The major and minor axes are perpendicular diameters of the ellipse. They define its shape and size. In the standard form \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\), the major axis is determined by the larger denominator (\(a^2\)), and the minor axis by the smaller one (\(b^2\)).For our ellipse \(\frac{x^{2}}{100} + \frac{y^{2}}{64} = 1\):

• The semi-major axis, \(a = 10\), indicates the major axis is horizontal, since 100 is larger than 64.
• The semi-minor axis, \(b = 8\), runs vertically.

The full length of each axis is twice the measure of its respective semi-axis:

• Major axis length: \(2a = 2 \times 10 = 20\)
• Minor axis length: \(2b = 2 \times 8 = 16\)

This 20-unit stretch horizontally and 16-unit stretch vertically gives the ellipse its distinctive elongated shape.