Problem 32 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the center, vertices, foci, and eccentricity of the ellipse, and sketch its graph. Use a graphing utility to verify your graph. $$\frac{(x+3)^{2}}{12}+\frac{(y-2)^{2}}{16}=1$$

Step-by-Step Solution

Answer

Center: (-3,2), Vertices: (-7,2), (1,2), Foci: (-5,2), (-1,2), Eccentricity: \(1/2\)

Step 1: Identify the center of the ellipse

The center of the ellipse \(h, k\) can be obtained from the equation. The equation \(\frac{(x+3)^{2}}{12}+\frac{(y-2)^{2}}{16}=1\) gives the center at (-3,2)

Step 2: Identify the semi-major axis (a) and semi-minor axis (b)

Comparing the denominators, we determine that 16 is greater than 12, hence the semi-major axis \(a\) is \(\sqrt{16} = 4\) and the semi-minor axis \(b\) is \(\sqrt{12} = 2\sqrt{3}\)

Step 3: Identify the vertices

The vertices of the ellipse are at a distance \(a\) from the center. Since \(a = 4\), the vertices will be located at (-3-4,2) = (-7,2) and (-3+4,2) = (1,2)

Step 4: Identify the foci

The foci of the ellipse are positioned at a distance \[c = \sqrt{a^2 - b^2}\] from the center. Here, c = \[\sqrt{(4)^2 - (2\sqrt{3})^2} = \sqrt{16 - 12} = 2\]. Hence, the foci are located at (-3-2,2) = (-5,2) and (-3+2,2) = (-1,2)

Step 5: Calculate the eccentricity

The eccentricity, e, of the ellipse is calculated by \[\sqrt{1 - \(\frac{b^2} {a^2}\)}\]. Substituting values, \[e = \sqrt{1 - \((2\sqrt{3})/4)^2\} = \sqrt{1 - 3/4} = 1/2\]

Step 6: Sketch the graph of the ellipse

With the center, vertices and foci, we can sketch the ellipse. Draw the horizontal and vertical axes. Mark the center at (-3,2). Draw the vertices at (-7,2) and (1,2) and the foci at (-5,2) and (-1,2). Draw the ellipse that passes through these points. Remember to verify the graph using a graphing utility.

Key Concepts

Center of Ellipse

The center of an ellipse is the point around which the ellipse is symmetrically formed. In the equation \( \frac{(x+3)^2}{12} + \frac{(y-2)^2}{16} = 1 \), the center is identified by looking at the values that make the perfect squares zero.
The general form equation of an ellipse \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \) gives the center as \((h, k)\).

• In our equation, \((h, k) = (-3, 2)\).
• The center is at point \((-3, 2)\), acting as the midpoint from which the ellipse expands equally in all directions based on the axes.

It's crucial to pinpoint this center, as it forms the basis for locating other key features.

Semi-Major Axis

The semi-major axis is one of the most important aspects of an ellipse. It represents the longest radius from the center to the edge of the ellipse.
In the given equation, we compare the denominators \(16\) and \(12\) to determine the length of the semi-major and semi-minor axes.

• The larger denominator \(16\) relates to the vertical semi-major axis because the y-variable is associated with it.
• The length of the semi-major axis \(a\) mathematically is \( \sqrt{16} = 4 \).

This length indicates that the ellipse stretches farther vertically than horizontally.

Vertices of Ellipse

Vertices are key endpoints of the semi-major axis. In our case, they lie along the vertical line through the center at \((-3, 2)\).
The vertices are found by adding and subtracting the semi-major axis length \(a\) from the center point.

• Start from the center \((-3, 2)\).
• Add and subtract \(a = 4\) along the vertical y-axis, giving the vertices: \((-3, 2 + 4) = (-3, 6)\) and \((-3, 2 - 4) = (-3, -2)\).

So, these points are the tips of the ellipse along its longest axis, essential for sketching its outline.

Foci of Ellipse

The foci are two unique points inside the ellipse, known for the property that the sum of distances from any point on the ellipse to the foci is constant.
To find the foci, we calculate \(c\) using the formula \(c = \sqrt{a^2 - b^2}\), where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes.

• Here, \(c = \sqrt{16 - 12} = 2\).
• Starting from the center \((-3, 2)\), move \(2\) units up and down along the y-axis: \((-3, 2 + 2) = (-3, 4)\) and \((-3, 2 - 2) = (-3, 0)\).

The foci play a vital role in defining the curve's shape.

Eccentricity of Ellipse

Eccentricity \(e\) measures how much an ellipse deviates from being a circle. It is a dimensionless number between 0 and 1. The formula for eccentricity is given by \(e = \sqrt{1 - \frac{b^2}{a^2}}\).
An eccentricity closer to 0 suggests a rounder shape, while closer to 1 indicates a more elongated form.

• In our situation, substitute the values: \(e = \sqrt{1 - \frac{12}{16}} = \sqrt{1 - 0.75} = \frac{1}{2}\).
• This \(e = \frac{1}{2}\) implies a moderately elongated ellipse.

Understanding eccentricity helps in visualizing the ellipse's overall form.