Problem 69 - A Graphical Approach to Precalculus with Limits

Question

Find an equation for each hyperbola. Vertices \((-3,0)\) and \((3,0) ;\) passing through \((6,1)\)

Step-by-Step Solution

Answer

The hyperbola equation is \(\frac{x^2}{9} - 3y^2 = 1\).

Step 1: Identify the Standard Form of the Hyperbola

Since the vertices are located at \((-3,0)\) and \((3,0)\), the hyperbola is horizontal. The standard form of a horizontal hyperbola centered at \((h, k)\) is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). With the given vertices, the center is \((0,0)\) and \(a = 3\). Hence, the equation is \(\frac{x^2}{9} - \frac{y^2}{b^2} = 1\).

Step 2: Use the Given Point to Find b²

Substitute the plug-in point \((6,1)\) into the equation \(\frac{x^2}{9} - \frac{y^2}{b^2} = 1\) to find \(b^2\). Thus \(\frac{6^2}{9} - \frac{1}{b^2} = 1\).

Step 3: Solve the Equation for b²

Calculate \(\frac{36}{9} = 4\). Then the equation becomes \(4 - \frac{1}{b^2} = 1\). So, \(-\frac{1}{b^2} = 1 - 4 = -3\), which means \(b^2 = \frac{1}{3}\).

Step 4: Write the Equation of the Hyperbola

Substituting \(b^2 = \frac{1}{3}\) into the equation gives \(\frac{x^2}{9} - \frac{y^2}{\frac{1}{3}} = 1\), which simplifies to \(\frac{x^2}{9} - 3y^2 = 1\).

Key Concepts

Standard Form of Hyperbola

Understanding the standard form of a hyperbola equation is vital to solving problems involving hyperbolas. The equation is slightly different based on whether the hyperbola is horizontal or vertical. In this problem, we are dealing with a horizontal hyperbola. The standard form for a horizontal hyperbola centered at \((h, k)\) is:\[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]Where:

• \((h, k)\) is the center of the hyperbola.
• \(a\) is the distance from the center to each vertex along the x-axis.
• \(b\) is the distance related to the asymptotes.

In this exercise, we've identified the center at \((0,0)\) with our vertices placed horizontally along the x-axis. This helps us choose the correct standard form and avoid confusion with a vertical hyperbola equation.

Vertices Calculation

Calculating the vertices of a hyperbola gives us important information about its size and placement. The vertices of the hyperbola are given at \((-3,0)\) and \((3,0)\). From this, we can extract the center and the value of \(a\):

• The center is the midpoint of the line segment joining the vertices. For our vertices, the midpoint can be calculated as:\[\left(\frac{-3+3}{2}, \frac{0+0}{2}\right) = (0,0)\]
• The value of \(a\) is half the distance between the vertices. The vertices are 6 units apart, so:\(a = \frac{6}{2} = 3\)

With this, we now know the center of the hyperbola and the \(a\) value necessary to plug into our standard form.

Center and Radius

In hyperbola problems, the center and radius play similar roles to those in circles, but with some differences. The center in a hyperbola is a point \((h, k)\).The radius concept is split into two distances:

• \(a\) - the distance from the center to each vertex on the x-axis for horizontal hyperbolas.
• \(b\) - changes the slope of the asymptotes and influences the shape of the hyperbola, although it is not a direct radius like in circles.

Understanding these values help us position our hyperbola correctly on the x-y plane. From our work so far, we've found the center is at \((0,0)\) with \(a = 3\).Let's not forget about \(b\), which we will solve shortly to finish our hyperbola equation.

Solving Equations

Solving the hyperbola's equation involves using the known point \((6,1)\) that lies on the hyperbola to find \(b^2\). Starting with our partial equation:\[\frac{x^2}{9} - \frac{y^2}{b^2} = 1\]Replace \(x\) and \(y\) with 6 and 1 respectively:

• Calculate: \(\frac{6^2}{9} = 4\)
• The equation simplifies to:\[4 - \frac{1}{b^2} = 1\]
• Solve for \(-\frac{1}{b^2}\):\(-\frac{1}{b^2} = 1 - 4 = -3\)
• Thus, \(b^2 = \frac{1}{3}\).

Integrating \(b^2\)into our equation provides us with the final form of the hyperbola's equation:\[\frac{x^2}{9} - 3y^2 = 1\] This process illustrates the importance of substituting known points into your equations to discover missing variables.