Problem 36 - A Graphical Approach to College Algebra

Question

Find an equation for each ellipse. Center \((-3,6) ;\) major axis vertical, with length \(10 ; c=2\)

Step-by-Step Solution

Answer

The equation is \(\frac{(x+3)^2}{21} + \frac{(y-6)^2}{25} = 1\).

Step 1: Understand the Characteristics of the Ellipse

The given ellipse has its center at the point \((-3, 6)\). It has a vertical major axis with a length of 10, which means that the semi-major axis \(a = \frac{10}{2} = 5\). The value \(c = 2\) is the distance from the center to each focus along the major axis.

Step 2: Calculate the Semi-minor Axis

For an ellipse, the relationship between the semi-major axis \(a\), the semi-minor axis \(b\), and the distance \(c\) is given by the equation \(c^2 = a^2 - b^2\). In this case, \(c=2\) and \(a=5\), so: \[ c^2 = a^2 - b^2 \]\[ 4 = 25 - b^2 \]\[ b^2 = 21 \]\[ b = \sqrt{21} \]

Step 3: Formulate the Equation of the Ellipse

Since the major axis is vertical, the standard form of the equation of the ellipse with center \((-3, 6)\) is:\[ \frac{(x+3)^2}{b^2} + \frac{(y-6)^2}{a^2} = 1 \]Substituting the known values \(b^2 = 21\) and \(a^2 = 25\), we get:\[ \frac{(x+3)^2}{21} + \frac{(y-6)^2}{25} = 1 \]

Key Concepts

Semi-major Axis

In an ellipse, the semi-major axis is half of the longest diameter. It's the distance from the center to the farthest point on the ellipse. When given a major axis length, you simply divide by two to find the semi-major axis.

For this exercise, the major axis is vertical, spanning a length of 10 units. Therefore, the semi-major axis is calculated as:

• \[ a = \frac{10}{2} = 5 \]

The semi-major axis helps in defining the ellipse's height and is crucial when writing the ellipse's equation. It determines the ellipse's stretch along the major axis. Always remember, this axis is associated with the variable having the larger denominator in the ellipse equation.

Semi-minor Axis

The semi-minor axis is the shortest radius of the ellipse. It is perpendicular to the semi-major axis and spans from the center to the edge of the ellipse.

For this calculation, we use the relationship between the semi-major axis \(a\), semi-minor axis \(b\), and the distance to each focus \(c\):

• \[ c^2 = a^2 - b^2 \]

Given \(c = 2\) and \(a = 5\), substituting these values in gives:

• \[ 4 = 25 - b^2 \]
• \[ b^2 = 21 \]
• \[ b = \sqrt{21} \]

The semi-minor axis impacts the width of the ellipse and corresponds to the smaller denominator in the ellipse equation.

Foci of an Ellipse

The foci of an ellipse are special points located along the major axis, equidistant from the center. They are not on the ellipse itself but inside it. The distance from the center to each focus is denoted by \(c\).

In this exercise, the value \(c = 2\) was given, which means:

• The foci are 2 units away from the center, located along the vertical axis at \((-3, 6+2)\) and \((-3, 6-2)\).
• Thus, the foci are at \((-3, 8)\) and \((-3, 4)\).

The foci are vital for understanding the shape and orientation of the ellipse. They define the eccentricity, influencing how stretched or circular the ellipse appears.