Problem 3 - 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. $$ 9 x^{2}+4 y^{2}+72 x-16 y+124=0 $$

Step-by-Step Solution

Answer

The given equation represents an ellipse.

Step 1: Group Terms

Start by rearranging the terms to group the x terms together and the y terms together: \[ 9(x^2 + 8x) + 4(y^2 - 4y) + 124 = 0 \] This step sets up the expression for completing the square.

Step 2: Complete the Square for x Terms

First, factor out the 9 from the x terms: \[ 9(x^2 + 8x + \, ) \]To complete the square, take half of the coefficient of x, 8, square it to get \(16\), and add & subtract inside the bracket: \[ 9((x+4)^2 - 16) \]Expand the completed square inside and simplify:\[ 9((x+4)^2 - 16) = 9(x+4)^2 - 144 \].

Step 3: Complete the Square for y Terms

Now factor out the 4 from the y terms: \[ 4(y^2 - 4y + \, ) \]To complete the square, take half of the coefficient of y, -4, square it to get \(4\), and add & subtract inside the bracket: \[ 4((y-2)^2 - 4) \]Expand the completed square inside and simplify: \[ 4((y-2)^2 - 4) = 4(y-2)^2 - 16 \].

Step 4: Substitute Completed Squares into the Equation

Substitute the squared terms back into the equation:\[ 9(x+4)^2 - 144 + 4(y-2)^2 - 16 + 124 = 0 \] Simplify by combining constant terms:\[ 9(x+4)^2 + 4(y-2)^2 = 36 \].

Step 5: Identify the Conic Section

Divide the whole equation by 36 to simplify:\[ \frac{9}{36}(x+4)^2 + \frac{4}{36}(y-2)^2 = 1 \]Simplify the fractions:\[ \frac{(x+4)^2}{4} + \frac{(y-2)^2}{9} = 1 \]This is the standard form of an ellipse equation: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \].

Key Concepts

Ellipse

The concept of an ellipse may initially seem complex, but it's quite manageable with a little understanding. An ellipse is a type of conic section that resembles an elongated circle. It's important in both mathematics and daily applications such as in the orbits of planets and satellites.
An ellipse is defined by two main axes:

• The major axis is the longest line that cuts through the center, extending to the edge of the ellipse.
• The minor axis is the shortest line that cuts through the center.

These axes intersect at the center of the ellipse. Its equation expresses this balance between x and y terms, resulting in an elliptical shape rather than a perfect circle. The above problem demonstrates how algebraic techniques can be used to identify and confirm the elliptical nature of a given equation.

Completing the Square

Completing the square is a powerful algebraic method used to transform quadratic equations into a form that is easier to understand and work with. In this case, it helps reframe the given equation into a recognizable format. Here's how it works:

• First, identify the quadratic terms and balance them by factoring out common coefficients.
• Next, tweak these terms to form perfect squares, a crucial step that reshapes our original terms into squared binomials.
• This involves taking half of the linear coefficient, squaring it, and then adding and subtracting this value within the expression.

For the equation in the exercise, completing the square was done separately for both x and y terms. Despite the fancy name, the process is fundamentally about creating squares, which makes everything easier to manipulate later on.

Standard Form of an Ellipse

Once you've completed the square, the next target is to transform the terms into the standard form of an ellipse. The standard equation for an ellipse is:\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]Here's how we get there from our original problem:

The goal is to arrange the equation such that the squares of binomials are visible, representing shifts from the origin. The constants, represented as \(a^2\) and \(b^2\), dictate the ellipse’s dimensions.

• \( h \) and \( k \) are the coordinates for the center of the ellipse.
• \( a \) and \( b \) are the semi-major and semi-minor axis lengths, respectively.

In the context of our problem, after simplification and dividing through, the coefficients in the equation mark the unmistakable layout of an ellipse. This step confirms the transformation into a geometrically significant form.

Algebraic Manipulation

Algebraic manipulation is crucial for reshaping and understanding equations. In this exercise, we rely on these skills to progress towards identifying the nature of the conic section. To better grasp this, consider these points:

• Rearrange terms: Begin by grouping like terms, making the equation easier to work with.
• Factor and balance: Factor out coefficients as needed to focus on completing the square smoothly.
• Simplify numerically: After modifying expressions, constant terms should be combined and reduced where possible.

Throughout this problem, manipulation is integral for adjusting and simplifying complicated expressions, leading effectively to the solution. In our case, careful arithmetic and methodical steps guide us to a neatly simplified equation representative of an ellipse. These skills are invaluable and extend beyond just conic sections. They are a staple of strong algebraic technique putting us in greater control over equations.