Problem 27 - College Algebra with Corequisite Support

Question

For the following exercises, find the foci for the given ellipses. \(\frac{(x+3)^{2}}{25}+\frac{(y+1)^{2}}{36}=1\)

Step-by-Step Solution

Answer

The foci are at \((-3, -1 + \sqrt{11})\) and \((-3, -1 - \sqrt{11})\).

Step 1: Identify the Standard Form

The given equation of the ellipse is \( \frac{(x+3)^{2}}{25} + \frac{(y+1)^{2}}{36} = 1 \). This matches the form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \) where \( (h, k) \) is the center, and \( a^2 = 25 \), \( b^2 = 36 \).

Step 2: Determine the Major and Minor Axes

Compare \( a^2 \) and \( b^2 \). Here, \( a^2 = 25 \) and \( b^2 = 36 \). Since \( b^2 > a^2 \), the major axis is vertical. Therefore, in this context \( a = 5 \) and \( b = 6 \).

Step 3: Calculate the Distance to the Foci

For ellipses, the distance from the center to each focus (\( c \)) is determined by \( c^2 = b^2 - a^2 \). Substitute the values \( b^2 = 36 \) and \( a^2 = 25 \) into the formula to get \( c^2 = 36 - 25 = 11 \). Therefore, \( c = \sqrt{11} \).

Step 4: Position the Foci

The foci of the ellipse, since it is vertical, are placed along the y-axis from the center. The center is \((-3, -1)\). This makes the foci \((-3, -1 + \sqrt{11})\) and \((-3, -1 - \sqrt{11})\).

Key Concepts

Standard Form of an Ellipse

Ellipses are fascinating geometric shapes that look like stretched circles. To work with ellipses, knowing their standard form is essential. An ellipse's equation appears in the format:

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

Here, \(h, k\) represents the center of the ellipse. \(a^2\) and \(b^2\) are the squared lengths of the semi-major and semi-minor axes. This standard form helps us understand the ellipse's orientation and dimensions. The values \(a\) and \(b\) dictate whether the ellipse is wider or taller. It's a foundational piece for finding other ellipse properties, like foci and axes lengths.
With our equation \( \frac{(x+3)^{2}}{25} + \frac{(y+1)^{2}}{36} = 1 \), you can see that the center is shifted to \(-3, -1\), changing the calculations slightly.

Foci of an Ellipse

Ellipses have two special points called foci (singular: focus). These points aren't always on the ellipse, but they define its shape. Imagine two thumbtacks and a tight rubber band pulling at them; the band’s path illustrates the ellipse. The sum of distances from any point on the ellipse to the two foci is constant.
The foci's coordinates differ based on the major axis's orientation. For our vertical ellipse, foci hover over and below the center. Given the center \(-3,-1\) and the vertical major axis, we place these foci with a precise vertical shift from the center. Understanding their placement is key in comprehending the ellipse's structure.

Major and Minor Axes

Within any ellipse, you'll find two axes: the major and minor. These line segments run through the ellipse center and determine the shape's width and height.

• The major axis is the longer one, extending along the ellipse's direction of maximum length. In our example, it runs vertically because \( b^2 > a^2 \) (since \( 36 > 25 \)). Thus the semi-major axis length is \( b = 6 \).
• The minor axis is shorter, perpendicular to the major axis. Here, its semi-minor length would be \( a = 5 \).

The axes' lengths directly affect the ellipse's stretch, with one axis longer mainly dictating whether the ellipse looks tall or wide.

Distance to Foci

Calculating the distance from the center to each focus, \(c\), informs the ellipse’s precise spread. To find \(c\), use the formula:

• \( c^2 = b^2 - a^2 \)

This equation gives a crucial hint on how foci position themselves. For our ellipse, substituting in our values:

• \( c^2 = 36 - 25 = 11 \)

Therefore, \( c = \sqrt{11} \). The value \(c\) informs how far from the center the foci sit along the major axis. Knowing this distance is crucial for ellipse analysis, providing insights into its balance between focal span and overall shape.