Problem 11 - College Algebra with Corequisite Support

Question

For the following exercises, determine which conic section is represented based on the given equation. \(2 x^{2}+3 y^{2}-8 x-12 y+2=0\)

Step-by-Step Solution

Answer

The equation represents an ellipse.

Step 1: Rewrite Equation in Standard Form

First, reorganize the given equation by grouping the terms associated with each variable: \(2x^{2} - 8x + 3y^{2} - 12y + 2 = 0\). We'll complete the square for both the \(x\) and \(y\) terms.

Step 2: Complete the Square for x-terms

Focus on the \(x\) terms: \(2x^{2} - 8x\). Factor out the 2: \(2(x^{2} - 4x)\). To complete the square inside the parentheses, take \(\frac{-4}{2} = -2\), square it to get 4, and add and subtract it inside: \(2(x^{2} - 4x + 4 - 4)\). The expression becomes \(2((x-2)^{2} - 4)\). Simplify to \(2(x-2)^2 - 8\).

Step 3: Complete the Square for y-terms

Now for the \(y\) terms: \(3y^{2} - 12y\). Factor out the 3: \(3(y^{2} - 4y)\). Take \(\frac{-4}{2} = -2\), square it to 4, and add/subtract it in the parentheses: \(3(y^{2} - 4y + 4 - 4)\), which simplifies to \(3((y-2)^{2} - 4)\). Simplify to \(3(y-2)^2 - 12\).

Step 4: Combine and Simplify the Equation

Substitute the completed squares back into the equation: \[2(x-2)^2 - 8 + 3(y-2)^2 - 12 + 2 = 0\]. Combine the constants: \(-8 - 12 + 2 = -18\). The equation is now \[2(x-2)^2 + 3(y-2)^2 = 18\].

Step 5: Identify the Conic Section

Divide the entire equation by 18 to simplify it into standard form: \[\frac{2(x-2)^2}{18} + \frac{3(y-2)^2}{18} = 1\]. Simplifying gives \[\frac{(x-2)^2}{9} + \frac{(y-2)^2}{6} = 1\]. This is the equation of an ellipse, since the equation is in the form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) with two positive terms.

Key Concepts

Equation of an Ellipse

Understanding the equation of an ellipse is crucial when working with conic sections. An ellipse is represented by the quadratic equation where the sum of the squared terms equals one:

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

In this formula, \((h, k)\) represents the center of the ellipse. It acts as the point from which the ellipse expands.
The variables \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively. Note, \(a\) is always associated with the larger denominator value. This determines whether the ellipse stretches horizontally or vertically.
If \(a^2 > b^2\), the ellipse is wider along the x-axis. Conversely, if \(b^2 > a^2\), it's taller along the y-axis.

When we say an equation is "of an ellipse," it means it follows this specific form with both terms having positive coefficients. Identifying these elements can greatly help in solving related exercises efficiently.

Completing the Square

Completing the square is a method to rewrite a quadratic equation into a perfect square trinomial plus or minus a constant. This process helps us transform equations to standard forms easily.

For a quadratic like \(ax^2 + bx\), first, factor out \(a\) if it isn't 1, yielding \(a(x^2 + \frac{b}{a}x)\). Then, take half of the coefficient of \(x\), square it, and add/subtract it within the parentheses:

• Find \(\frac{b}{2a}\)
• Square it, resulting in \(\left(\frac{b}{2a}\right)^2\)
• Add and subtract this value inside the parentheses to form a square trinomial

In our ellipse example, the \(x\) terms \(2x^2 - 8x\) were completed by adding and subtracting 4 inside the equation to form \((x-2)^2\). This simplification made it possible to put the problem in standard form, leading to the equation of an ellipse.

Applying this technique can simplify solving conics by enabling easier identification of their general forms.

Standard Form of Conics

The standard form of conics is essential for classifying and understanding different types of conic sections. Each conic section has its own standard equation:

For an ellipse, like the one in our exercise, the form is:

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

The terms are set so the equation equals 1, and both fractions involve positive coefficients. This differs for different conics:

• For a circle: \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{a^2} = 1\) where \(a = b\)
• For a hyperbola: The equation sets one term as negative, e.g., \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)
• For a parabola: Only one squared term exists, as in \((x-h)^2 = 4p(y-k)\)

Recognizing these forms simplifies the process of identifying the type of conic section represented by an equation. Once we've completed the square for our ellipse, converting to standard form helped us confirm its type by seeing it matches the ellipse form precisely.