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Chapter 8: Problem 69

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

Short Answer

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

Step by step solution

01

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 \(\frac{x^2}{9} - \frac{y^2}{b^2} = 1\).
02

Use the Given Point to Find b虏

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 \(\frac{6^2}{9} - \frac{1}{b^2} = 1\).
03

Solve the Equation for b虏

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 \(b^2 = \frac{1}{3}\).
04

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Use substitution to solve the nonlinear system of equations in three variables. Note that solutions are ondered triples. $$\begin{aligned} 2 x^{2}+y^{2}+3 z^{2} &=3 \\ 2 x+y-z &=1 \\ x+y &=0 \end{aligned}$$

Sound Detection Microphones are placed at points \((-c, 0)\) and \((c, 0) .\) An explosion occurs at point \(P(x, y)\) having positive \(x\) -coordinate. The sound is detected at the closer microphone \(t\) seconds before being detected at the farther microphone. Assume that sound travels at a speed of 330 meters per second, and show that \(P\) must be on the hyperbola $$ \frac{x^{2}}{330^{2} t^{2}}-\frac{y^{2}}{4 c^{2}-330^{2} t^{2}}=\frac{1}{4} $$ (GRAPH CANT COPY)

CONCEPT CHECK The graph of the rational function \(y=\frac{1}{x}\) is a hyperbola that is rotated. Experiment with a graphing calculator to determine the vertices of its graph.

Graph the solution set of each system of inequalities by hand. $$\begin{aligned} &\frac{x^{2}}{4}+\frac{y^{2}}{9} \leq 1\\\ &\frac{y^{2}}{4}-\frac{x^{2}}{9} \leq 1 \end{aligned}$$

Graph the solution set of each system of inequalities by hand. $$\begin{aligned} &\frac{(x-2)^{2}}{36}+\frac{(y+2)^{2}}{25} \leq 1\\\ &\frac{(x+1)^{2}}{9}+\frac{(y-3)^{2}}{25} \leq 1 \end{aligned}$$
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