Problem 30 - College Algebra

Question

For the following exercises, find the foci for the given ellipses. $$ x^{2}+4 y^{2}+4 x+8 y=1 $$

Step-by-Step Solution

Answer

The foci are at \((-2 \pm \frac{3\sqrt{3}}{2}, -1)\).

Step 1: Rearrange the ellipse equation

Start by rearranging the given equation: \[ x^{2} + 4y^{2} + 4x + 8y = 1 \]Group the x-terms together and the y-terms together:\[ (x^2 + 4x) + 4(y^2 + 2y) = 1 \]

Step 2: Complete the square for x-terms

Complete the square for the expression \(x^2 + 4x\). 1. Take half of the coefficient of \(x\), which is \(4\), divide by \(2\) to get \(2\).2. Square \(2\) to get \(4\).3. Add and subtract \(4\) inside the parentheses:\[ (x^2 + 4x + 4 - 4) \]\[ (x+2)^2 - 4 \]

Step 3: Complete the square for y-terms

Complete the square for the expression \(y^2 + 2y\).1. Take half of the coefficient of \(y\), which is \(2\), divide by \(2\) to get \(1\).2. Square \(1\) to get \(1\).3. Add and subtract \(1\) inside the parentheses:\[ 4(y^2 + 2y + 1 - 1) \]\[ 4((y+1)^2 - 1) \]

Step 4: Rewrite the completed squares

Rewrite the rearranged equation using the completed squares:\[ (x+2)^2 - 4 + 4((y+1)^2 - 1) = 1 \]Simplify the expression:\[ (x+2)^2 - 4 + 4(y+1)^2 - 4 = 1 \]Combine and simplify constants:\[ (x+2)^2 + 4(y+1)^2 = 9 \]

Step 5: Divide to form ellipse standard equation

Divide the entire equation by \(9\) to get the ellipse in standard form:\[ \frac{(x+2)^2}{9} + \frac{(y+1)^2}{rac{9}{4}} = 1 \]The standard form of an ellipse is compared to:\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \] where \(a^2 = 9\) and \(b^2 = \frac{9}{4}\).

Step 6: Identify the center and semi-major/minor axes

From the ellipse equation, the center \(h = -2, k = -1\), \(a^2 = 9\), and \(b^2 = \frac{9}{4}\). Notice that \(a > b\), indicating a horizontal ellipse:- Semi-major axis \(a = 3\)- Semi-minor axis \(b = \frac{3}{2}\)

Step 7: Calculate the foci distance

The formula for the distance to the foci, \(c\), is given by:\[ c = \sqrt{a^2 - b^2} \]Substitute the values \(a^2 = 9\) and \(b^2 = \frac{9}{4}\).\[ c = \sqrt{9 - \frac{9}{4}} = \sqrt{\frac{36}{4} - \frac{9}{4}} = \sqrt{\frac{27}{4}} = \frac{3\sqrt{3}}{2} \]

Step 8: Determine the foci

The foci are located along the x-axis (horizontal) because this is a horizontal ellipse.The center is \(-2, -1\), so:- Foci = \((-2 \pm \frac{3\sqrt{3}}{2}, -1)\)Which gives:- Foci 1: \((-2 + \frac{3\sqrt{3}}{2}, -1)\)- Foci 2: \((-2 - \frac{3\sqrt{3}}{2}, -1)\)

Key Concepts

Completing the Square

Completing the square is a mathematical technique used to transform a quadratic expression, making it easier to work with in various applications, such as converting an equation to a different form. In our exercise, we apply this to both the x- and y-terms of the given ellipse equation.

For the x-terms,

• we look at the expression \(x^2 + 4x\).
• Half of 4 is 2, and squaring it gives us 4.
• We add and subtract 4 in the expression: \((x+2)^2 - 4\).

Similarly, for the y-terms:

• start with \(y^2 + 2y\).
• Half of 2 is 1, squared is 1.
• Add and subtract 1: \((y+1)^2 - 1\).

These steps effectively "complete the square," making it easier to identify features of the ellipse like its center.

Ellipse Standard Form

The standard form of an ellipse equation is pivotal for understanding its key properties. By following a process of completing the square on the equation expressed as \(x^2 + 4y^2 + 4x + 8y = 1\), we successfully reformat the equation into its standard form.

Once you have completed the square, the equation should resemble:\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \] with variables indicating:

• \((h, k)\) as the ellipse's center coordinates.
• \(a^2\) and \(b^2\) as the denominators representing the squares of the semi-major and semi-minor axes.

For our exercise, we equate terms leading to the transformed equation:\[ \frac{(x+2)^2}{9} + \frac{(y+1)^2}{\frac{9}{4}} = 1 \]This shows us a horizontal ellipse with the longer axis along the x-direction.

Foci Calculation

Calculating the foci of an ellipse is essential in revealing the nature of this geometric figure. The foci are always located on the major axis of the ellipse, and determining their position involves understanding the distance, \(c\), from the center.

This distance is found using the formula:\[ c = \sqrt{a^2 - b^2} \] where \(a\) and \(b\) are derived from the equation in standard form.

• Substitute \(a^2 = 9\) and \(b^2 = \frac{9}{4}\).
• \(c = \sqrt{9 - \frac{9}{4}} = \sqrt{\frac{27}{4}} = \frac{3\sqrt{3}}{2}\).

These calculations show that the foci are positioned equidistant from the center points along the x-axis. They follow the coordinates:

• Foci 1: \((-2 + \frac{3\sqrt{3}}{2}, -1)\)
• Foci 2: \((-2 - \frac{3\sqrt{3}}{2}, -1)\)

Center of Ellipse

Determining the center of an ellipse is crucial, allowing us to explore further properties like the axes lengths and foci placement. From the equation in its standard form,\[ \frac{(x+2)^2}{9} + \frac{(y+1)^2}{\frac{9}{4}} = 1 \] we can identify the center as \((-2, -1)\).

This transformation follows from observing the expressions within the parentheses:

• \((x+2)^2\) indicates \(x\) is centered at \(-2\).
• \((y+1)^2\) shows \(y\) is centered at \(-1\).

The center is essential not only for plotting the ellipse correctly but also for determining its orientation and the calculation results for foci and axes.