Problem 50 - College Algebra

Question

For the following exercises, use the given information about the graph of each ellipse to determine its equation. Center \((3,5) ;\) vertex \((3,11) ;\) one focus: \((3,5+4 \sqrt{2})\)

Step-by-Step Solution

Answer

The ellipse equation is \(\frac{{(x-3)^2}}{4} + \frac{{(y-5)^2}}{36} = 1\).

Step 1: Identify the center and vertex

The center of the ellipse is given as \((3, 5)\) and a vertex as \((3, 11)\). This indicates that the major axis is vertical since both the center and the vertex share the same x-coordinate.

Step 2: Calculate the semi-major axis

The distance from the center \((3, 5)\) to the vertex \((3, 11)\) is the length of the semi-major axis (\(a\)). This distance is \(11 - 5 = 6\). Hence, \(a = 6\).

Step 3: Determine the distance to the focus

The focus is given as \((3, 5 + 4\sqrt{2})\). The distance from the center \((3, 5)\) to the focus is the value of \(c\). Calculate this as \(5 + 4\sqrt{2} - 5 = 4\sqrt{2}\), so \(c = 4\sqrt{2}\).

Step 4: Calculate the semi-minor axis

Use the relationship \(c^2 = a^2 - b^2\) to find \(b\). Given \(c = 4\sqrt{2}\) and \(a = 6\), compute:\[ (4\sqrt{2})^2 = 6^2 - b^2 \ 32 = 36 - b^2 \ b^2 = 4 \] Thus, \(b = 2\).

Step 5: Write the equation of the ellipse

Since the major axis is vertical, the standard form of the ellipse's equation is:\[ \frac{{(x-h)^2}}{b^2} + \frac{{(y-k)^2}}{a^2} = 1 \]Substitute \(h = 3\), \(k = 5\), \(a = 6\), and \(b = 2\):\[ \frac{{(x-3)^2}}{4} + \frac{{(y-5)^2}}{36} = 1 \]

Key Concepts

Semi-Major Axis

The semi-major axis is one of the most important parts of an ellipse. It's the longest radius that stretches from the center to the edge of the ellipse, and it tells us a lot about the ellipse's size and orientation. In the exercise given, with the center at \((3,5)\) and a vertex at \((3,11)\), we can find the length of this axis by calculating how far one vertex is from the center.
Simply put, this is the distance between \((3,5)\) and \((3,11)\), which equals 6. So, the semi-major axis \(a\) equals 6.
The semi-major axis is crucial because it dictates the width of the ellipse in that direction. In our case, a vertical major axis suggests that the ellipse is taller than it is wide.

Semi-Minor Axis

The semi-minor axis lies perpendicular to the semi-major axis. It's shorter and provides the "width" of the ellipse from the center to its edge along the minor axis.To find the length of the semi-minor axis \(b\), we use the relationship between the axes and the distance to the foci:

• The formula is \(a^2 = b^2 + c^2\).
• Here, \(a = 6\) and the distance to the focus \(c = 4\sqrt{2}\).

By plugging these into the formula, - We calculate \(b^2\) as \(a^2 - c^2\). - With \((6)^2\) being 36 and \((4\sqrt{2})^2\) being 32, we get: - \(b^2 = 36 - 32 = 4\). Thus, \(b = 2\). This shorter segment confirms that the ellipse shape is stretched more vertically.

Standard Form of Ellipse

The equation of an ellipse describes its shape and orientation in a single mathematical expression. For our ellipse, since the major axis is vertical, the standard form looks like this:
\(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\)
Here, \((h,k)\) is the center of the ellipse. For our exercise, this is \((3,5)\). The values of \(a\) and \(b\) are the semi-major and semi-minor axes we've calculated: \(a = 6\) and \(b = 2\).

• Substitute \((h, k) = (3, 5)\).
• Use \(b^2 = 4\) and \(a^2 = 36\).

The standard form of the ellipse equation, thus, becomes:
\[\frac{(x-3)^2}{4} + \frac{(y-5)^2}{36} = 1\]
This equation fully encapsulates the shape and position of our ellipse in a two-dimensional plane.

Distance to Focus

The distance to the focus \(c\) is another key feature of an ellipse. This helps to determine the shape's eccentricity—how stretched or circular it is.Foci are special points inside the ellipse, not located on the axes, but crucial for its definition. In the exercise, one focus is given as \((3, 5 + 4\sqrt{2})\).

• To find this distance \(c\), we measure how far the focus \((3, 5+4\sqrt{2})\) is from the center \((3,5)\).
• This computes as \(5 + 4\sqrt{2} - 5 = 4\sqrt{2}\).

Hence, \(c = 4\sqrt{2}\). This value is essential when using the relation \(a^2 = b^2 + c^2\) to find the semi-minor axis. Focus distance offers insights not only into size but also into the ellipse's focal properties.