Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

Name the conic or limiting form represented by the given equation. Usually you will need to use the process of completing the square. $$ 3 x^{2}+3 y^{2}-6 x+12 y+60=0 $$

Step-by-Step Solution

Answer

The equation represents an empty set, indicating no real conic shape.

Step 1: Group and Rearrange Terms

First, rearrange the given equation by grouping it into terms with variables and constant terms. The equation is: \[ 3x^2 - 6x + 3y^2 + 12y + 60 = 0 \]Group the terms: \[ (3x^2 - 6x) + (3y^2 + 12y) = -60 \]

Step 2: Factor Out Constants

Factor out the common factor of 3 from both groups of variable terms:\[ 3(x^2 - 2x) + 3(y^2 + 4y) = -60 \]Then, divide the entire equation by 3 to simplify:\[ (x^2 - 2x) + (y^2 + 4y) = -20 \]

Step 3: Complete the Square for x-terms

To complete the square for the x-terms, take half of the coefficient of x, square it, and add it inside the square. The expression is:\[ (x^2 - 2x) \]Half of \(-2\) is \(-1\), and squaring it gives 1:\[ (x^2 - 2x + 1) - 1 \]

Step 4: Complete the Square for y-terms

Similarly, complete the square for the y-terms using the expression:\[ (y^2 + 4y) \]Half of 4 is 2, and squaring it gives 4:\[ (y^2 + 4y + 4) - 4 \]

Step 5: Rewrite Equation with Completed Squares

Insert the completed squares back into the equation:\[ (x^2 - 2x + 1) + (y^2 + 4y + 4) = -20 + 1 + 4 \]This simplifies to:\[ (x - 1)^2 + (y + 2)^2 = -15 \]

Step 6: Analyze the Resulting Equation

The resulting equation \[ (x - 1)^2 + (y + 2)^2 = -15 \] indicates that it represents a conic section.Since the right side of the equation is negative, this configuration represents an empty set in the real plane, indicating no real solutions (thus not physically forming a conic shape).

Key Concepts

Completing the Square

Completing the square is a useful algebraic technique, especially when dealing with quadratic equations. It transforms a quadratic expression into a perfect square trinomial.

Here's how it works:

• Start with a quadratic in the form of \( ax^2 + bx + c \).
• Focus on the \( ax^2 + bx \) part and move the constant \( c \) to the other side of the equation if necessary.
• Take the coefficient of \( x \), halve it, and then square it. This number completes your square.
  For instance, in the expression \( x^2 - 2x \), halve \(-2\) to get \(-1\), and square it to make \(1\).
• Add and subtract this squared number within the expression to create a perfect square trinomial.
  This ensures our expression takes the form \((x - 1)^2\) here.

Completing the square is particularly useful for solving quadratic equations or transforming equations into a form that reveals the properties of conic sections.

Quadratic Equations

Quadratic equations are polynomials of degree 2, and they take the standard form \( ax^2 + bx + c = 0 \).

They are pivotal in algebra and mathematics! Here's why:

• They graph as parabolas, which can open either upwards or downwards depending on the sign of \( a \).
• The solutions to these equations are the values where the parabola intersects the x-axis; these solutions can be found using various methods, like factoring, completing the square, or the quadratic formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \)!
• Quadratics are also foundational in understanding conic sections, as they are embedded in equations representing circles, ellipses, hyperbolas, and parabolas.

Quadratic equations are everywhere in math, linking various concepts together, including the understanding of the geometry of curves.

Real Solutions

Real solutions refer to the values of a variable that satisfy an equation and result in real numbers.

In the context of quadratic equations:

• If the discriminant \( b^2 - 4ac \) is positive, the quadratic equation has two distinct real solutions.
• If the discriminant is zero, there's exactly one real solution, meaning the parabola touches the x-axis at a point known as the vertex.
• When the discriminant is negative, there are no real solutions; the solutions are instead complex or imaginary, which means the parabola does not intersect the x-axis.

In our specific problem, completing the square led to a result that represented no real solutions. The fact that the terms summed to a negative number indicates that the equation does not correspond to a conic section visible in the real plane. Instead, it results in an empty set, making it crucial to analyze such results when interpreting equations in geometry.