Problem 30 - College Algebra with Corequisite Support

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 of the ellipse are at \((-2 + \frac{3\sqrt{3}}{2}, -1)\) and \((-2 - \frac{3\sqrt{3}}{2}, -1)\).

Step 1: Rearrange the Equation

To find the foci of the ellipse described by the equation \(x^2 + 4y^2 + 4x + 8y = 1\), we first rearrange the equation to be in the standard form of an ellipse. First, we'll complete the square for the \(x\) and \(y\) terms.

Step 2: Complete the Square for \(x\)

First, let's focus on completing the square for the \(x\) terms. We have \(x^2 + 4x\). To complete the square, we take half of the coefficient of \(x\), which is 4, divide by 2 to get 2, and square it to get 4. We add and subtract 4 inside the equation:\(x^2 + 4x = (x+2)^2 - 4\).

Step 3: Complete the Square for \(y\)

Next, we complete the square for the \(y\) terms. We have \(4y^2 + 8y\). Factor out the 4:\(4(y^2 + 2y)\). Now complete the square for \(y^2 + 2y\):Half of 2 is 1, and squaring it gives 1. Add and subtract 1 inside:\(y^2 + 2y = (y+1)^2 - 1\).Now the expression becomes:\(4((y+1)^2 - 1) = 4(y+1)^2 - 4\).

Step 4: Simplify the Equation

Substitute the completed square forms back into the original equation:\((x+2)^2 - 4 + 4((y+1)^2 - 1) = 1\)Simplifying this, we get:\((x+2)^2 + 4(y+1)^2 = 9\).Divide the entire equation by 9 to get it into standard form:\(\frac{(x+2)^2}{9} + \frac{(y+1)^2}{9/4} = 1\).

Step 5: Identify the Parameters of the Ellipse

In the standard form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), here \(h = -2\), \(k = -1\), \(a^2 = 9\) (meaning \(a = 3\)), and \(b^2 = \frac{9}{4}\) (meaning \(b = \frac{3}{2}\)).

Step 6: Calculate the Distance of the Foci

The distance \(c\) of the foci from the center is given by the equation \(c = \sqrt{a^2 - b^2}\).Calculate \(c\):\(c = \sqrt{9 - \frac{9}{4}} = \sqrt{\frac{36}{4} - \frac{9}{4}} = \sqrt{\frac{27}{4}} = \frac{3\sqrt{3}}{2}\).

Step 7: Determine the Location of the Foci

Since \(a > b\), the ellipse is oriented horizontally, and the foci are located at \((h \pm c, k)\).Plugging in the values, the foci are located at:\((-2 \pm \frac{3\sqrt{3}}{2}, -1)\).Thus, the foci are: \((-2 + \frac{3\sqrt{3}}{2}, -1)\) and \((-2 - \frac{3\sqrt{3}}{2}, -1)\).

Key Concepts

Completing the Square

To transform the equation of an ellipse into a more workable form for solving, we use a method known as "completing the square." This technique helps simplify both the

• linear and square terms
• into a perfect square trinomial

by balancing the equation. For instance, take the expression involving the variable x: \[x^2 + 4x\]. First, identify the coefficient of x, which is 4. Divide this coefficient by 2 and then square it, resulting in \[(4/2)^2 = 4.\]We then write \[x^2 + 4x = (x+2)^2 - 4\]. For the y terms \[4y^2 + 8y\],we start by factoring out the 4, leaving \[(y^2 + 2y)\]. Repeat the process by taking half of the 2, squaring it to obtain 1, and then expressing it as \[y^2 + 2y = (y+1)^2 - 1\].Factoring is critical in maintaining the balance of an equation while transforming its structure, effectively aiding in solving the problem at hand.

Standard Form of an Ellipse

An ellipse equation takes a simplified form called the "standard form." This form allows us to easily identify key features such as the center, vertices, and foci. The general form of an ellipse is: \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1,\]where:

• \((h, k)\) is the center of the ellipse,
• \(a\) is the semi-major axis (if \(a > b\)), and
• \(b\) is the semi-minor axis (if \(b < a\)).

For our ellipse, after completing the square, the equation becomes \[\frac{(x+2)^2}{9} + \frac{(y+1)^2}{\frac{9}{4}} = 1.\]This indicates that the ellipse is centered at \((-2, -1)\),with the horizontal semi-major axis length \(a = 3\) and the vertical semi-minor axis length \(b = \frac{3}{2}\). This clarity and visual orientation allows us to understand the geometric structure of the ellipse at a glance.

Distance Formula for Foci

To find the foci of an ellipse, we utilize the distance formula for foci. This formula is essential to determine the precise measurement for the location of the ellipse's foci from its center. If the ellipse has a horizontal major axis (meaning \(a > b\)), the distance \(c\) from the center to each focus is given by: \[c = \sqrt{a^2 - b^2}.\]For the equation:

• \(a = 3\)
• \(b = \frac{3}{2}\)

Putting these values into the formula provides \[c = \sqrt{9 - \frac{9}{4}} = \frac{3\sqrt{3}}{2}.\]The foci are then located at a distance of \(\frac{3\sqrt{3}}{2}\) units horizontally from the center at \((-2, -1)\).Thus, the precise locations of the foci for this ellipse are \((-2 + \frac{3\sqrt{3}}{2}, -1)\) and \((-2 - \frac{3\sqrt{3}}{2}, -1)\). Understanding this relationship allows for a deeper comprehension of how ellipses are constructed and analyzed mathematically.