Problem 59 - Algebra and Trigonometry Real Mathematics, Real People

Question

Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. \(16 x^{2}-9 y^{2}+32 x+54 y-209=0\)

Step-by-Step Solution

Answer

The given equation is a hyperbola.

Step 1: Group and reorder terms

Group the x and y terms together and then move the constant term to the right side of the equation. \n(16x^{2}+32x) - (9y^{2}-54y) = 209

Step 2: Complete the square

For each group, use the formula \((b/2a)^{2}\) to calculate the constant you need to add and subtract to complete the square. The equation becomes: \n16(x^{2}+2x+1) -9(y^{2}-6y+9) = 209 + 16 - 81.

Step 3: Simplify

Simplify the equation and present it in the standard form. \n16(x+1)^{2} - 9(y-3)^{2} = 144.

Step 4: Identify the curve

The standard form of the equation is of the form \((x-h/w)^{2} - (y-k/v)^{2} = 1\), where w and v > 0 . This form is recognizable as a hyperbola.

Key Concepts

Hyperbola

A hyperbola is a type of conic section formed when a plane intersects both nappes of a double cone. Understanding hyperbolas is crucial because they represent shapes and equations encountered in various scientific fields. A hyperbola consists of two disjoint curves, known as branches, which mirror each other. It can be expressed in standard mathematical form as:

• Horizontal axis: \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)
• Vertical axis: \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\)

Here, \(h,k\) identifies the hyperbola's center, and \(a,b\) determine its spread along the axes. This general equation differs from those of ellipses, whose terms are effectively added rather than subtracted. This subtraction of squared terms creates the unique, unbounded structure of a hyperbola. Recognizing a hyperbola often starts with rewriting mathematical expressions to match the standard form using methods such as completing the square.

Completing the Square

Completing the square is a mathematical technique often used to transform quadratic equations into a form that makes solving them clearer. It's essential for working with conic sections like hyperbolas. When applied to the original equation, it helped reveal the hyperbola's standard form. Here's how it works in stages:First, we reorder and group terms associated with the same variable. For example, the given equation was organized to group all terms involving \(x\) and \(y\):\[16x^2 + 32x - 9y^2 + 54y = 209\]Next, we add and subtract specific constants to "complete" each variable's square inside each group. This transforms a quadratic into a perfect square trinomial. Using the formula \((b/2a)^2\), we determine what constant to add for each group so it becomes a perfect square trinomial:

• \((x^2 + 2x + 1)\) for the \(x\) terms
• \((y^2 - 6y + 9)\) for the \(y\) terms

Adding and subtracting these constants within the equation ensures the expression's integrity is maintained, while its form becomes more manageable. Completing the square transforms the problem into something easier to solve or reduce to known forms.

Quadratic Equations

Quadratic equations are a fundamental concept underpinning various fields of mathematics and science. They're particularly important when dealing with conic sections, as they often represent the core component of these shapes' algebraic definitions. In the exercise, the equation contains the quadratic elements \(x^2\) and \(y^2\), prompting the need to employ strategies like completing the square for simplification.A typical quadratic equation takes the form:\[ax^2 + bx + c = 0\]where \(a\), \(b\), and \(c\) are coefficients. The solution to quadratics can be found using various methods:

• Factoring
• The Quadratic Formula: \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)
• Completing the Square, as done in this example

Transforming a quadratic equation, especially when part of a complex equation like one representing a hyperbola, is vital. It enables us to re-express it in forms that reveal the path it describes, such as the two branches of a hyperbola, while also making it easier to solve or analyze. Not only does this deep understanding empower problem-solving, but it also strengthens the foundation for further mathematical exploration, including calculus and more advanced algebra.