Problem 45 - College Algebra

Question

For the following exercises, find the foci for the given ellipses. $$ 4 x^{2}+16 x+4 y^{2}+16 y+16=0 $$

Step-by-Step Solution

Answer

The foci are at (-2, -2).

Step 1: Reorganize the equation

First, we need to reorganize the given equation into a standard form. The original equation is: \[4x^2 + 16x + 4y^2 + 16y + 16 = 0.\] Divide the whole equation by 4 to simplify:\[x^2 + 4x + y^2 + 4y + 4 = 0.\] Next, move the constant to the right side:\[x^2 + 4x + y^2 + 4y = -4.\]

Step 2: Complete the square

To convert the equation into the standard form of an ellipse, complete the square for both the \(x\) and \(y\) terms.For \(x\):- Take half of the coefficient of \(x\) (which is 4), square it: \(\left(\frac{4}{2}\right)^2 = 4\).- Add and subtract 4 inside the equation: \((x^2 + 4x + 4)\).For \(y\):- Take half of the coefficient of \(y\) (which is 4), square it: \(\left(\frac{4}{2}\right)^2 = 4\).- Add and subtract 4 inside the equation: \((y^2 + 4y + 4)\).The equation becomes:\[(x+2)^2 + (y+2)^2 = 4.\]

Step 3: Identify the standard form

Identify the standard form of the ellipse from the equation obtained after completing the square:\[(x+2)^2 + (y+2)^2 = 4.\]Recognize that this is of the form \[(x-h)^2 + (y-k)^2 = r^2\], where \((h, k)\) is the center, and \(r\) is the radius of a circle. Since the coefficients of \(x\) and \(y\) are equal and the equation represents a circle, the foci lie at the same point as the center in this special case of an ellipse.

Step 4: Determine foci of the ellipse

For an ellipse described by \[(x-h)^2 + (y-k)^2 = r^2\],the foci are usually found using \(c\) where \(c^2 = |a^2 - b^2|\) (applicable in case of non-equal axes). In this case \((h, k) = (-2, -2)\) and \(r^2 = 4\), so since it's a circle (a special case of an ellipse with coinciding major axes), the foci are the same as the center: \((-2, -2)\).

Key Concepts

Completing the Square

Completing the square is a helpful technique in algebra used to transform quadratic equations into a form that is easier to analyze or solve. This method is especially useful in writing the equation of a conic section, such as an ellipse, in its standard form. Here’s a simple way to understand it:

• Start by identifying the quadratic terms in your equation. You’ll need to focus on the terms with variables squared, such as \(x^2\) and \(y^2\).
• To complete the square for a term like \(x^2 + 4x\), take half of the linear coefficient (the number in front of \(x\)), which is 4 in this case. Half of 4 is 2.
• Square the result. Therefore, \(2^2 = 4\). Now, add and subtract this number (4) to complement your square: \((x^2 + 4x + 4 - 4)\).
• Organize this into a perfect square: \((x+2)^2-4\).

Repeat these steps for the \(y\) variable. Completing the square transforms the original equation into a tidier, more analyzable form. This process is crucial when you're working with conic sections and helps you find important features like the center and radius of a circle or ellipse.

Standard Form Equation

The standard form of a conic section’s equation allows us to clearly identify its key properties, such as center, axes, and foci. For circles and ellipses, this is particularly useful.
For a circle, the standard form is:

• \((x-h)^2 + (y-k)^2 = r^2\),

where \((h, k)\) is the center of the circle, and \(r\) is its radius.
For an ellipse, the standard forms vary slightly depending on the orientation:

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

In these forms, \((h,k)\) is the center, \(a\) and \(b\) are related to the distances from the center to the ellipse, along the major and minor axes respectively. Having the equation in a standard form makes it much easier to interpret the geometry of the shape and find crucial features like foci using the relationships set by the geometric definitions of circles and ellipses.

Circle as Special Ellipse

Often thought about separately, a circle is technically a special case of an ellipse where both axes are the same length. This makes learning about ellipses very practical since one concept strengthens your understanding of the other. In an ellipse:

• The major axis is the longest diameter.
• The minor axis is the shortest diameter.

However, in the case of a circle, these axes are equal, effectively simplifying the equation. This means that the foci, which usually vary in an ellipse according to its stretched dimensions, become one single point—the center of the circle. This implies:

• Foci coincide with the circle’s center.
• The circle's equation \((x-h)^2 + (y-k)^2 = r^2\) is essentially the simplest form of an ellipse equation.

Understanding circles as special ellipses not only helps in finding foci but also deepens the conceptual grasp of geometric shapes and their equations in algebra.