Problem 13 - 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. \(\frac{y^{2}}{4}-\frac{x^{2}}{81}=1\)

Step-by-Step Solution

Answer

The hyperbola has vertices at (0, 2) and (0, -2), foci at (0, \(\sqrt{85}\)) and (0, -\(\sqrt{85}\)), and asymptotes at \(y = \pm \frac{2}{9}x\).

Step 1: Identify the Conic Type

The given equation \( \frac{y^2}{4} - \frac{x^2}{81} = 1 \) is the standard form of a hyperbola. A hyperbola in this form suggests that it is a vertical hyperbola because the \( y^2 \) term is positive.

Step 2: Confirm Standard Form

Verify if the given equation \( \frac{y^2}{4} - \frac{x^2}{81} = 1 \) is in standard form for a vertical hyperbola. The standard form is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), which matches, confirming the hyperbola is already in the standard form.

Step 3: Identify Vertices

For a vertical hyperbola \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the vertices are at \( (0, \pm a) \). Here, \( a^2 = 4 \) so \( a = 2 \). Therefore, the vertices are at \( (0, 2) \) and \( (0, -2) \).

Step 4: Identify Foci

The foci of a vertical hyperbola are at \( (0, \pm c) \) where \( c = \sqrt{a^2 + b^2} \). Here, \( c = \sqrt{4 + 81} = \sqrt{85} \). Thus, the foci are located at \( (0, \sqrt{85}) \) and \( (0, -\sqrt{85}) \).

Step 5: Write Equations of Asymptotes

For a vertical hyperbola \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the asymptotes are given by the equations \( y = \pm \frac{a}{b}x \). With \( a = 2 \) and \( b = 9 \), the equations are \( y = \pm \frac{2}{9}x \).

Key Concepts

Vertices of Hyperbola

The vertices of a hyperbola are key points that give us a clear view of the hyperbola's "open ends." In essence, they represent the points where the hyperbola intersects its transverse axis, which is the axis connecting these vertices.
For a vertical hyperbola, like in the equation given:

• Standard form: \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)
• The vertices are located at \( (0, \pm a) \)

This is because in a vertical hyperbola, the axis along the \( y \)-axis is the one stretching out and forming the shape.
From the given exercise, we have \( a^2 = 4 \), hence \( a = 2 \). Therefore, the vertices of the hyperbola are at \( (0, 2) \) and \( (0, -2) \).
Recognizing and locating the vertices helps sketch the hyperbola accurately and provides foundational understanding when dealing with these types of conic sections.

Foci of Hyperbola

Foci (plural of focus) in a hyperbola serve as important points reflecting its nature and definition. They help describe the path followed by points in the form of a difference of distances.
Each hyperbola has two foci, and for a vertical hyperbola, they are positioned along the same axis as the vertices—along the \( y \)-axis.
To find the coordinates of the foci, we use the formula:

• \( c = \sqrt{a^2 + b^2} \)
• The foci are located at \( (0, \pm c) \)

In our exercise:

• \( a^2 = 4 \), and \( b^2 = 81 \)
• \( c = \sqrt{4 + 81} = \sqrt{85} \)

Thus, the foci are at the coordinates \( (0, \sqrt{85}) \) and \( (0, -\sqrt{85}) \).
The distance between each focus and any point on the hyperbola remains constant when taking the difference between the distances to both foci, which is a beautiful property of hyperbolas that stems from these specific locations.

Asymptotes of Hyperbola

Asymptotes in a hyperbola are straight lines that the hyperbola approaches but never touches as it extends to infinity. They essentially provide a boundary that guides the shape and direction of the hyperbola.
The equations of the asymptotes for a vertical hyperbola in standard form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) are:

• \( y = \pm \frac{a}{b}x \)

Using our exercise values:

• \( a = 2 \)
• \( b = 9 \)

We derive the asymptote equations:

• \( y = \frac{2}{9}x \)
• \( y = -\frac{2}{9}x \)

These lines intersect at the center of the hyperbola and lay the framework within which the hyperbola exists.
Visualizing these asymptotes aids in developing an understanding of the hyperbola's plot and behavior. They are instrumental in constraining the hyperbola's path as its branches extend outward.