Problem 36 - A Graphical Approach to Precalculus with Limits

Question

Determine the type of conic section represented by each equation, and graph it, provided a graph exists. $$3 x^{2}+12 x+3 y^{2}=0$$

Step-by-Step Solution

Answer

The equation describes a circle centered at (-2, 0) with radius 2.

Step 1: Identify the Squared Terms

The equation given is \(3x^2 + 12x + 3y^2 = 0\). We observe that it has two squared terms: \(3x^2\) and \(3y^2\). Both variables, \(x\) and \(y\), are squared and have the same coefficient when factored out, suggesting it might represent a circle or an ellipse.

Step 2: Simplify the Equation

First, simplify the given equation by factoring out the common coefficient from the squared terms. The equation becomes \(3(x^2 + 4x + y^2) = 0\). Dividing both sides by 3 gives \(x^2 + 4x + y^2 = 0\).

Step 3: Complete the Square

To complete the square for the \(x\) terms, consider \(x^2 + 4x\). Half of 4 is 2, and 2 squared is 4. We add and subtract 4 inside the equation: \(x^2 + 4x + 4 - 4 + y^2 = 0\). This simplifies to \((x + 2)^2 + y^2 = 4\).

Step 4: Identify the Conic Section

The completed square form \((x + 2)^2 + y^2 = 4\) is that of a circle, as it fits the standard form \((x - h)^2 + (y - k)^2 = r^2\), where \(h = -2\), \(k = 0\), and \(r = 2\).

Step 5: Determine the Effect of the Equation

Because \((x + 2)^2 + y^2 = 4\) describes a circle centered at \((-2, 0)\) with a radius of 2, the equation indeed corresponds to a circle. This verifies that the conic section is a circle, with the graph existing in the coordinate plane.

Key Concepts

Equation Simplification

Equation simplification is a crucial technique in mathematics that helps to make complex problems more manageable. A simplified equation is generally easier to analyze and solve. Let's simplify the equation step-by-step.

In our exercise, the given equation was:

• \(3x^2 + 12x + 3y^2 = 0\)

The first step in simplification involves identifying and factoring out common factors in the equation. Here, we see that both \(3x^2\) and \(3y^2\) share the common factor of 3. By factoring 3 out of the equation, we can rewrite it as:

• \(3(x^2 + 4x + y^2) = 0\)

Next, by dividing every term by 3, we further simplify the equation to:

• \(x^2 + 4x + y^2 = 0\)

This step both reduces the complexity and prepares the equation for the next step, which is completing the square. Simplification not only makes manipulation easier but also helps in recognizing the forms of equations, such as recognizing conic sections.

Completing the Square

Completing the square is a method often used for converting a quadratic equation into a form that involves a perfect square trinomial.
This technique is helpful when you need to solve for variables or when you aim to recognize the equation's conic section form.

In this particular exercise, we look at the terms involving \(x\) in the simplified equation \(x^2 + 4x + y^2 = 0\). The goal is to transform \(x^2 + 4x\) into a perfect square trinomial.Here's how it’s done:

• Take half of the coefficient of \(x\) (which is 4), resulting in 2.
• Square it, yielding 4.
• Add and subtract this square within the equation to maintain equality.
  • \(x^2 + 4x + 4 - 4 + y^2 = 0\)

Upon completing these steps, we form

• \((x + 2)^2 + y^2 = 4\)

This form is now much more recognizable as a standard conic equation.
Completing the square transitions the equation into a form that is easier to graph and analyze.

Circle Graph

Understanding circle graphs is essential when identifying and sketching conic sections. The standard equation of a circle in the coordinate plane is

• \((x - h)^2 + (y - k)^2 = r^2\)

where

• \((h, k)\) denotes the center
• \(r\) represents the radius

In our completed form

• \((x + 2)^2 + y^2 = 4\)

we recognize:

• The circle’s center is at \((-2, 0)\).
• The radius is 2, since \(4 = r^2\) leads to \(r = 2\).

Knowing the circle's center and radius enables you to easily sketch its graph. Start by plotting the center on the coordinate plane, then use the radius to draw the circle surrounding that point.

This visualization can assist in providing insight into the equation’s geometric properties. When identifying a graph as a circle, these characteristics can facilitate a better understanding of the underlying relationships in the coordinate plane.