Problem 81 - A Graphical Approach to College Algebra

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: Rearrange the Equation

Start by rearranging the given equation to group the terms with the same variables. The given equation is \(x^2 - 6x - 2y^2 + 7 = 0\). Move constants to the other side: \(x^2 - 6x - 2y^2 = -7\).

Step 2: Complete the Square for x

Focus on completing the square for the \(x\)-terms. Take \(x^2 - 6x\). The coefficient of \(x\) is -6. Divide it by 2 to get -3, and then square it to get 9. Add and subtract 9 to maintain equality: \((x^2 - 6x + 9) - 9\).

Step 3: Include Completed Square in the Equation

Add the completed square to the equation. Now, \( (x - 3)^2 - 9 - 2y^2 = -7 \). Simplify to get \((x - 3)^2 - 2y^2 = 2 \).

Step 4: Bring to Standard Form

Divide the entire equation by 2 so that the equation represents a hyperbola. The standard form is \( \frac{(x - 3)^2}{2} - \frac{y^2}{1} = 1 \).

Step 5: Identify the Center

The center \((h, k)\) of the hyperbola from the standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) is \((3, 0)\).

Step 6: Calculate the Vertices

In the standard form \(\frac{(x-3)^2}{2} - \frac{y^2}{1} = 1\), the vertices are \((h \pm a, k)\). Here, \(a = \sqrt{2}\), so the vertices are \((3 \pm \sqrt{2}, 0)\).

Key Concepts

Standard Form of a Hyperbola

A hyperbola in its standard form is written as \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) when it opens horizontally. This form helps us to identify key features like its center, vertices, and asymptotes. Here:

• \( (h, k) \) is the center of the hyperbola.
• \( a \) and \( b \) are real numbers which determine the shape and size of the hyperbola.

In our problem, after completing the square and rearranging terms, the equation becomes \( \frac{(x - 3)^2}{2} - \frac{y^2}{1} = 1 \). This equation now represents a horizontally opening hyperbola in standard form, making subsequent steps like finding the center and vertices straightforward. Remember, the key to getting to this form is accurately rearranging and completing the square on the relevant terms.

Center of Hyperbola

The center point \((h, k)\) defines the midpoint from which the hyperbola symmetrically extends. For a hyperbola written in standard form, this center is crucial as all other features like vertices and asymptotes are measured with respect to it.
In our example, the center is found by identifying \(h\) and \(k\) from the equation \( \frac{(x - 3)^2}{2} - \frac{y^2}{1} = 1 \). Here, by comparing to the general form \( \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \), we identify the center of this hyperbola as \((3, 0)\).
This center indicates the point around which the entire hyperbola is organized, providing a reference point for plotting and analysis.

Vertices of Hyperbola

Vertices are key points on a hyperbola that lie on its transverse axis. For a horizontally opening hyperbola, the vertices are located at \((h \pm a, k)\). Here, \(a\) is derived from the term \(\frac{(x-h)^2}{a^2}\) in the standard form equation.
In our specific case, the standard form is \( \frac{(x - 3)^2}{2} - \frac{y^2}{1} = 1 \), which gives us \(a^2 = 2\). Therefore, \(a = \sqrt{2}\).
So, the vertices of the hyperbola are found at \((3 \pm \sqrt{2}, 0)\). These vertices are indicative of the widest part of the hyperbola, marking where the curve comes closest together along the horizontal section.

Completing the Square

Completing the square is a method used to simplify a quadratic expression, enabling easier manipulation and conversion to standard forms like those used for conic sections. For a quadratic \(x^2 - 6x\), we:

• Take the coefficient of \(x\) (which is -6), divide by 2 to get -3, and square it to add and subtract that value (9 in this case).
• Thus, \(x^2 - 6x\) becomes \((x - 3)^2 - 9\).

In our exercise, completing the square allowed us to convert \(x^2 - 6x - 2y^2 = -7\) into a form that's easier to manipulate: \((x - 3)^2 - 2y^2 = 2\).
This technique is crucial for deriving the standard form of the hyperbola, providing a systematic approach to rewrite and solve quadratic expressions.