Problem 27

Question

Find an equation for the hyperbola that has its center at the origin and satisfies the given conditions. Vertices \(V(\pm 3,0), \quad\) asymptotes \(y=\pm 2 x\)

Step-by-Step Solution

Verified
Answer
The equation of the hyperbola is \( \frac{x^2}{9} - \frac{y^2}{36} = 1 \).
1Step 1: Understand the Hyperbola's Orientation
The vertices are given at \((\pm 3,0)\), indicating that the hyperbola is centered at the origin and opens horizontally. This suggests the standard form of the equation is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
2Step 2: Identify the Value of 'a'
The vertices of the hyperbola are at \((\pm a, 0)\). Given \((\pm 3, 0)\) as the vertices, it follows that \(a = 3\). Thus, \(a^2 = 9\).
3Step 3: Relate Asymptotes to 'a' and 'b'
For a hyperbola centered at the origin with the equation form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the slopes of the asymptotes are \( \pm \frac{b}{a} \). Given the asymptotes \(y = \pm 2x\), the slope is \(2\), therefore, \(\frac{b}{a} = 2\).
4Step 4: Solve for 'b'
Substitute \(a = 3\) into \(\frac{b}{a} = 2\), which gives \(\frac{b}{3} = 2\). Solving for \(b\), we get \(b = 6\). Hence, \(b^2 = 36\).
5Step 5: Write the Equation of the Hyperbola
Now that we have \(a^2 = 9\) and \(b^2 = 36\), substitute these into the standard form to get the equation of the hyperbola: \( \frac{x^2}{9} - \frac{y^2}{36} = 1 \).

Key Concepts

Vertices of HyperbolaHyperbola AsymptotesHyperbola CenterStandard Form of Hyperbola
Vertices of Hyperbola
Vertices of a hyperbola provide important information about its size and orientation. In the problem, the vertices are located at \((\pm 3,0)\). This tells us a couple of things:
  • The hyperbola is centered at the origin because the vertices are symmetrical around \(x = 0\).
  • The hyperbola opens horizontally along the x-axis since the y-coordinate of both vertices is 0.

The distance from the center to each vertex is known as \(a\). Here, \(a = 3\), which is the distance from the center (at 0,0) to each vertex. So, the vertices are integral in determining the orientation and shape of the hyperbola.
Hyperbola Asymptotes
Asymptotes in a hyperbola are straight lines that the hyperbola approaches but never actually reaches. They provide a direction to the branches of the hyperbola and essentially 'guide' their shape.
Given that the asymptotes are \(y=\pm 2x\), we know:
  • The slopes of these lines are \(\pm 2\).
  • This slope tells us how steeply the hyperbola's branches will open.
With the slope \(\pm \frac{b}{a} = \pm 2\), we can find the value of \(b\) since we already know \(a = 3\). By solving \(\frac{b}{3} = 2\), we find \(b = 6\). The asymptotes help us deduce the complete shape and equation of a hyperbola, running through its center at the origin.
Hyperbola Center
The center of a hyperbola is an essential reference point. It is the midpoint between the vertices and also lies on the intersection of its asymptotes. For this problem, the center is given as the origin, \((0,0)\). Knowing the center helps us place the vertices and sketch the hyperbola correctly.
  • The center provides symmetry to the hyperbola; it divides the hyperbola into mirror images.
  • The asymptotes intersect at the center, dictating the paths along which the hyperbola's arms diverge.

Understanding where the hyperbola is centered helps us position the equation's variables \(x\) and \(y\) correctly in its standard equation.
Standard Form of Hyperbola
The standard form of a hyperbola's equation depends on its orientation. Since our hyperbola opens horizontally, the equation takes the form of \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]Here, \(a\) and \(b\) are distances that relate to the hyperbola's shape and orientation.
  • \(a^2\) is the denominator in the \(x^2\) term since our hyperbola is horizontal.
  • \(b^2\) is the denominator in the \(y^2\) term.

For this particular problem, \(a = 3\) meaning \(a^2 = 9\), and \(b = 6\) which gives \(b^2 = 36\). Substituting into the standard form yields:\[ \frac{x^2}{9} - \frac{y^2}{36} = 1 \]This standard form allows us to easily analyze the hyperbola's properties and symmetries, making it central to understanding hyperbolas in algebra.