Problem 42 - College Algebra

Question

For the following exercises, find the foci for the given ellipses. $$ 64 x^{2}+128 x+9 y^{2}-72 y-368=0 $$

Step-by-Step Solution

Answer

The foci are at \((-1, 4 + \sqrt{55})\) and \((-1, 4 - \sqrt{55})\).

Step 1: Write the equation in standard form for an ellipse

The standard form of an ellipse equation is \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). First, we need to complete the square for both the \(x\) and \(y\) terms to rewrite the given equation. Start by grouping the \(x\) and \(y\) terms: \[ 64(x^2 + 2x) + 9(y^2 - 8y) = 368 \]

Step 2: Complete the square for the x terms

To complete the square for \(x\), take the coefficient of \(x\), divide by 2, and square it. Here, the coefficient is 2. \[ (2/2)^2 = 1 \]Add and subtract 1 inside the parenthesis: \[ 64((x^2 + 2x + 1) - 1) = 64((x+1)^2 - 1) \]Substitute back into the equation:\[ 64((x+1)^2 - 1) + 9(y^2 - 8y) = 368 \]

Step 3: Complete the square for the y terms

To complete the square for \(y\), take the coefficient of \(y\), divide by 2, and square it. Here, the coefficient is -8. \[ (-8/2)^2 = 16 \]Add and subtract 16 inside the parenthesis: \[ 9((y^2 - 8y + 16) - 16) = 9((y-4)^2 - 16) \]Substitute back into the equation:\[ 64((x+1)^2 - 1) + 9((y-4)^2 - 16) = 368 \]

Step 4: Simplify and rewrite in standard form

Distribute and simplify the completed square terms:\[ 64(x+1)^2 - 64 + 9(y-4)^2 - 144 = 368 \]Combine and move constant terms:\[ 64(x+1)^2 + 9(y-4)^2 = 576 \]Divide the entire equation by 576 to set it equal to 1:\[ \frac{(x+1)^2}{9} + \frac{(y-4)^2}{64} = 1 \]

Step 5: Identify the semi-major and semi-minor axes

In the equation \( \frac{(x+1)^2}{9} + \frac{(y-4)^2}{64} = 1 \), compare with the standard form to identify \(a^2 = 64\) (vertical major axis) and \(b^2 = 9\) (horizontal minor axis). Thus, \(a = 8\) and \(b = 3\).

Step 6: Calculate the distance to the foci

For ellipses, the square of the distance from the center to the foci \(c\) is given by \(c^2 = a^2 - b^2\). Calculate:\[ c^2 = 64 - 9 = 55 \]Thus, \(c = \sqrt{55} \).

Step 7: Determine the coordinates of the foci

The center of the ellipse is \((-1, 4)\). Since it is a vertical ellipse, the foci are located along the \(y\)-axis. Therefore, the coordinates of the foci are:\[ (-1, 4 + \sqrt{55}) \text{ and } (-1, 4 - \sqrt{55}) \]

Key Concepts

Completing the Square

"Completing the square" is a technique used to convert a quadratic expression into a perfect square trinomial plus a constant. This method will help us rewrite the ellipse equation in its standard form.

• Begin by grouping the terms with the same variable together.
• Complete the square separately for both the \(x\) and \(y\) terms.

For instance, if we have \(64(x^2 + 2x) + 9(y^2 - 8y)\), we need to focus on each group.

For the \(x\) terms:

• Take the coefficient of \(x\), which is 2, divide by 2, and square it to get 1.
• Add and subtract this result inside the parenthesis.
• The expression becomes \((x+1)^2 - 1\).
• Multiply by 64 to maintain equation balance.

For the \(y\) terms:

• Take the coefficient of \(y\), which is -8, divide by 2, and square it to get 16.
• Add and subtract 16 inside the parenthesis.
• The expression becomes \((y-4)^2 - 16\).
• Multiply by 9 similarly.

This method helps transition the quadratic terms into a setup that is easier to manipulate into the standard form of an ellipse.

Standard Form of an Ellipse

An ellipse's equation in standard form is crucial for analyzing its properties. The standard form is given by:
\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]
This form reveals the center, axes, and orientation of the ellipse.

• The point \((h, k)\) serves as the ellipse's center.
• \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively.

Depending on whether \(a > b\) or \(a < b\), the ellipse will have different orientations.
In our example, once we've completed the square, the equation is converted into:
\[ \frac{(x+1)^2}{9} + \frac{(y-4)^2}{64} = 1 \]
Here, we can see that:

• The center is at \((-1, 4)\).
• \(a^2 = 64\), meaning the semi-major axis is in the y-direction and \(a = 8\).
• \(b^2 = 9\), with the semi-minor axis perpendicular to the major axis and \(b = 3\).

Understanding the standard form helps in identifying key ellipse features easily.

Ellipse Equation

The starting point for tackling any ellipse problem is the "ellipse equation". An ellipse equation in its general quadratic form might not immediately present its nature due to mixed and combined terms.
To convert it to a more informative structure, we employ techniques like completing the square.
This conversion reveals:

• Key ellipse attributes and symmetry.
• Direction of stretching based on axes lengths.

For example, the original ellipse equation was:
\[ 64 x^{2}+128 x+9 y^{2}-72 y-368=0 \]
By restructuring this equation through the completing the square process, we transition into the more insightful:
\[ \frac{(x+1)^2}{9} + \frac{(y-4)^2}{64} = 1 \]
This form allows us to easily spot and interpret the ellipse's properties, including its center and axes measurements.

Distance from Center to Foci

The foci of an ellipse are pivotal in understanding its geometric properties, especially how elongated it might be. The "distance from center to foci" determines where these foci lie along the major axis.
Calculating this distance involves:

• Knowing the values of \(a^2\) and \(b^2\) from the equation.

For ellipses, this distance \(c\) is calculated with the equation:
\(c^2 = a^2 - b^2\). Here, \(a\) is the length of the semi-major axis, and \(b\) is the length of the semi-minor axis.
In the example,

• \(a^2 = 64\) and \(b^2 = 9\).

This means:
\(c^2 = 64 - 9 = 55\), giving us \(c = \sqrt{55}\).
With the center at \((-1, 4)\) and a vertical major axis, the foci positions can be calculated as:
\((-1, 4 + \sqrt{55})\) and \((-1, 4 - \sqrt{55})\).
These positions emphasize the ellipse's symmetry and geometric form.