Problem 81 - A Graphical Approach to Precalculus with Limits

Question

Write the equation in standard form for a hyperbola centered at ( \(h, k\) ). Identify the center and vertices. $$x^{2}-6 x-2 y^{2}+7=0$$

Step-by-Step Solution

Answer

Center: (3, 0). Vertices: (3 ± √2, 0).

Step 1: Reorganize and Simplify

Start by reorganizing the terms of the equation: \(x^2 - 6x - 2y^2 + 7 = 0\). Move constant to the other side: \(x^2 - 6x - 2y^2 = -7\).

Step 2: Complete the Square for x

To complete the square for the \(x\) terms: take half of the coefficient of \(x\), which is -3, square it to get 9, and add this inside the expression. Rewrite it as \((x-3)^2 -9\). Now the equation becomes: \((x-3)^2 - 9 - 2y^2 = -7\).

Step 3: Rearrange Equation and Balance

Add 9 to both sides to balance the equation: \((x-3)^2 - 2y^2 = 2\).

Step 4: Factor the Equation

Factor by recognizing that this is a form of a hyperbola. Divide through by 2 to standardize: \(\frac{(x-3)^2}{2} - \frac{y^2}{1} = 1\).

Step 5: Identify the Center, Axes, and Vertices

From the equation \(\frac{(x-3)^2}{2} - \frac{y^2}{1} = 1\), the center of the hyperbola is \((h, k) = (3, 0)\). For the hyperbola in the form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), \(a^2 = 2\) and \(b^2 = 1\). The vertices are \((h \pm a, k) = (3 \pm \sqrt{2}, 0)\).

Key Concepts

Standard Form Equation

The standard form equation of a hyperbola helps us to easily identify its key features. For a hyperbola centered at the origin, the standard form equation typically looks like this:

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

These forms help identify whether the hyperbola opens horizontally or vertically. The variables \(h\) and \(k\) denote the center of the hyperbola, \((h, k)\). Likewise, \(a\) and \(b\) reflect the distances from the center to each vertex along the hyperbola's axes.
In our solved problem, we started with the equation \(x^2 - 6x - 2y^2 + 7 = 0\). By reorganizing and simplifying, we eventually reached the standard form of a hyperbola: \( \frac{(x-3)^2}{2} - \frac{y^2}{1} = 1 \). This indicates that our hyperbola is centered at \((3, 0)\) and opens horizontally, as \(x\) is associated with the positive term. Understanding and transforming equations into their standard form makes analyzing the properties of conic sections straightforward.

Completing the Square

Completing the square is a powerful algebraic technique used to simplify quadratic expressions and equations. This method involves creating a perfect square trinomial in order to aid in solving or transforming equations. To complete the square:

• Take the coefficient of \(x\) (or \(y\) when needed), divide it by 2, and then square it.
• Add that squared term inside the expression, and balance it by adding or subtracting the same value on the other side of the equation.

In our exercise, we applied completing the square to the term \(x^2 - 6x\). Half of \(-6\) is \(-3\), and squaring \(-3\) gives us \(9\). This allows us to rewrite \(x^2 - 6x \) as \((x-3)^2 - 9\), simplifying the equation to a format that makes identifying the hyperbola's standard form easier.
This method is essential for converting many conic section equations into their standard forms, as it provides clarity on diagnosing the structure and characteristics of the conic section involved.

Conic Sections

Conic sections are the curves obtained at the intersection of a plane with a double-napped cone. They include circles, ellipses, parabolas, and hyperbolas. Each type has its distinct equation and properties. A hyperbola, specifically, is a type of conic section with characteristics comparable to two mirrored images.
The general form of a hyperbola is similar to other conic sections, but its unique feature lies in its subtraction form within the equation. That's how you distinguish it from ellipses, where the terms are added. This negative structure indicates that it opens outward, either along the x-direction or the y-direction.

• The hyperbola in the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) primarily opens in the x-direction.
• Conversely, \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \) indicates it opens in the y-direction.

Understanding conic sections, especially hyperbolas, helps unveil many natural and architectural patterns, ranging from planetary orbits to gradients of architectural structures. Recognizing the type through its standard equation form assists in visualizing and understanding the geometrical and physical properties represented therein. This is crucial not only for tasks in geometry but also for real-world data modeling.