Problem 15 - A Graphical Approach to College Algebra

Question

The equation of a conic section is given in a familiar form. Identify the type of graph (if any) that each equation has, without actually graphing. See the summary chart in this section. Do not use a calculator. $$\frac{(x+3)^{2}}{16}+\frac{(y-2)^{2}}{16}=1$$

Step-by-Step Solution

Answer

The graph is a circle.

Step 1: Identify the Form of Equation

The given equation is \( \frac{(x+3)^{2}}{16} + \frac{(y-2)^{2}}{16} = 1 \). This resembles the standard form of the equation for an ellipse: \( \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1 \), where \((h, k)\) is the center.

Step 2: Determine Parameters

Comparing \( \frac{(x+3)^2}{16} + \frac{(y-2)^2}{16} = 1 \) with the standard form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), we determine that \( h = -3 \), \( k = 2 \), and \( a^2 = b^2 = 16 \). This indicates it is an ellipse.

Step 3: Check the Relation of a and b

In an ellipse, the values \( a^2 \) and \( b^2 \) represent the denominators in the standard form, where \( a^2 = b^2 \) means that \( a = b \). This particular scenario forms a circle because both axes have the same length.

Step 4: Identify the Type of Conic Section

Since the equation \( \frac{(x+3)^2}{16} + \frac{(y-2)^2}{16} = 1 \) gives \( a = b \), it confirms this conic section is not just any ellipse but is specifically a circle.

Key Concepts

Ellipse

An ellipse is a type of conic section that appears as an elongated circle. Think of an ellipse as a "stretched-out" circle. It has two focal points, and the sum of the distances from these focal points to any point on the ellipse is constant. When we see an equation resembling the format \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), we're looking at the standard form of an ellipse.
Here are key features to look for when determining if you're dealing with an ellipse:

• The coefficients of \( (x-h)^2 \) and \( (y-k)^2 \) are different, except when the ellipse is actually a circle.
• The denominators, \( a^2 \) and \( b^2 \), tell us about the lengths of the axes. If \( a^2 eq b^2 \), the ellipse is stretched more in one direction.
• The center of the ellipse is at \((h, k)\), shifting from the origin based on these values.

Understanding these core traits allows us to identify an ellipse correctly when analyzing conic equations, even without graphing.

Circle

A circle is a special type of ellipse where the major and minor axes are of equal length. This makes it not only a simple shape but also the simplest of the conic sections. When you see an equation in the form \( \frac{(x-h)^2}{r^2} + \frac{(y-k)^2}{r^2} = 1 \), you are looking at the equation of a centered circle in its standard position.
Key characteristics of a circle include:

• The value of \( a^2 \) is equal to \( b^2 \), which implies \( a = b \).
• The center of the circle is given by \((h, k)\), creating a perfect symmetry around this point.
• The radius \( r \) is given by the common denominator (\( a^2 = b^2 = r^2 \)).

Circles are especially straightforward to identify within conic equations because of their symmetry and uniformity in axis lengths.

Equation Identification

Identifying whether a conic equation represents an ellipse, circle, or other conic sections is quite foundational in algebra. Instead of graphing, we rely on the form of the equation itself. Recognizing these types by their standard forms is crucial for proper interpretation.
Here are useful identification tips:

• If both terms in the equation have equal coefficients and equal denominators, it is likely a circle, assuming the equation is structured like an ellipse.
• If the denominators are different, then we are usually dealing with an ellipse when the equation is in the form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \).
• Being able to compare \( a^2 \) and \( b^2 \) without needing to graph ensures you can quickly identify the conic section type.

Practice with identifying these based on the equations will make determining conic sections much easier in your studies.