Problem 12 - College Algebra with Corequisite Support

Question

For the following exercises, write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes. \(\frac{x^{2}}{100}-\frac{y^{2}}{9}=1\)

Step-by-Step Solution

Answer

Vertices: (10,0) and (-10,0); Foci: (\(\sqrt{109} \),0) and (-\(\sqrt{109} \),0); Asymptotes: \( y = \pm \frac{3}{10}x \).

Step 1: Identify Hyperbola Type

The given equation is \( \frac{x^{2}}{100} - \frac{y^{2}}{9} = 1 \). Since the \( x^2 \) term is positive and comes before the \( y^2 \) term, this is a horizontal hyperbola of the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \).

Step 2: Determine Center, a, and b

For a horizontal hyperbola, the standard form is \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). From \( \frac{x^2}{100} - \frac{y^2}{9} = 1 \), we identify \( h = 0 \), \( k = 0 \), \( a^2 = 100 \), and \( b^2 = 9 \). Thus, \( a = 10 \) and \( b = 3 \). The center is at \((0,0)\).

Step 3: Find Vertices

For a horizontal hyperbola, the vertices are located at \((h \pm a, k)\). Substituting the values gives us \((0 \pm 10, 0)\) or vertices at \((10, 0)\) and \((-10, 0)\).

Step 4: Calculate Foci

The distance to the foci from the center is calculated using \( c = \sqrt{a^2 + b^2} \). Thus, \( c = \sqrt{100 + 9} = \sqrt{109} \). The foci are located at \((h \pm c, k)\), so they are \((\sqrt{109}, 0)\) and \((-\sqrt{109}, 0)\).

Step 5: Write Equations of Asymptotes

The equations of the asymptotes for a horizontal hyperbola are \( y = \pm \frac{b}{a} (x - h) + k \). With \( b = 3 \) and \( a = 10 \), the asymptotes are \( y = \frac{3}{10}x \) and \( y = -\frac{3}{10}x \).

Key Concepts

Standard Form of a Hyperbola

Whenever you see an equation of a hyperbola, it's usually written in its standard form. The standard form helps us easily identify important features of the hyperbola, like whether it's horizontal or vertical, as well as other characteristics like its vertices and foci.
For a hyperbola, the standard equation is typically written as:

• Horizontal hyperbola: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)
• Vertical hyperbola: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)

Here, \((h, k)\) is the center of the hyperbola.- The values \(a^2\) and \(b^2\) come from the denominators of their respective fractions.- If the \(x^2\) term appears first and is positive, you have a horizontal hyperbola, just like in the problem: \( \frac{x^{2}}{100} - \frac{y^{2}}{9} = 1 \).Understanding the standard form is critical, so you can easily navigate finding other features, such as vertices, foci, and asymptotes.

Vertices and Foci of a Hyperbola

Vertices and foci are essential parts of a hyperbola that give us more insight into its shape. Let's unpack these terms a bit:

• Vertices: For a horizontal hyperbola, vertices are the points located along the x-axis from the center. To find these, use \((h \pm a, k)\) where \(a\) is calculated from \(a^2\), the denominator of the \(x^2\) term. In our case, we have vertices at \((10, 0)\) and \((-10, 0)\).
• Foci: The foci of a hyperbola are points located along the transverse axis, which help define the curvature of the hyperbola. For our exercise, the foci are determined by \((h \pm c, k)\) using the formula \(c = \sqrt{a^2 + b^2}\). So we have the foci at \((\sqrt{109}, 0)\) and \((-\sqrt{109}, 0)\).

The vertices and foci help outline the overall direction and shape of the hyperbola, providing clues on how wide or narrow it might be.

Equations of Asymptotes for a Hyperbola

Asymptotes are imaginary lines that a hyperbola approaches but never actually touches. They guide the overall shape and direction of the hyperbola's branches. Let's simplify how to find their equations.
For a horizontal hyperbola, like the one in our exercise problem, we use the formula for the asymptotes:

• \( y = \pm \frac{b}{a}(x-h) + k \)

Where \(b\) and \(a\) are derived as the roots of the denominators of the \(y^2\) and \(x^2\) terms, respectively.
In our exercise:- \(a = 10\) and \(b = 3\), meaning the asymptotes will be \(y = \frac{3}{10}x\) and \(y = -\frac{3}{10}x\).These equations help us identify how the hyperbola stretches and where it approximates these lines, giving a visual interpretation of its reach and the direction for each "arm" of the graph.