Problem 13 - College Algebra

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

Standard form is correct. Vertices: \((0, \pm 2)\); Foci: \((0, \pm \sqrt{85})\); Asymptotes: \(y = \pm \frac{2}{9}x\).

Step 1: Identify Hyperbola Type

The given equation is \( \frac{y^2}{4} - \frac{x^2}{81} = 1 \). Since the \( y^2 \) term is positive, this is a vertical hyperbola of the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).

Step 2: Determine Values of a and b

In the standard form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), we identify \( a^2 = 4 \) and \( b^2 = 81 \). Solving these gives \( a = 2 \) and \( b = 9 \).

Step 3: Identify Vertices

For a vertical hyperbola, the vertices are at \((0, \pm a)\). Substitute \( a = 2 \) to find the vertices: \((0, 2)\) and \((0, -2)\).

Step 4: Calculate Foci

For hyperbolas, the distance to the foci is given by \( c = \sqrt{a^2 + b^2} \). Calculate \( c = \sqrt{4 + 81} = \sqrt{85} \). Thus, the foci are at \((0, \pm \sqrt{85})\).

Step 5: Write Equations of Asymptotes

The asymptotes for a vertical hyperbola are given by \( y = \pm \frac{a}{b}x \). Substitute the values of \( a \) and \( b \) to find the equations of the asymptotes: \( y = \pm \frac{2}{9}x \).

Key Concepts

Standard form of hyperbola equation

Hyperbolas are fascinating geometric shapes that come from conic sections. Their equations may look complex, but they follow a pattern that can be broken down with simplicity. The standard form of a hyperbola depends on its orientation and can greatly help in identifying its properties. For a hyperbola centered at the origin, the equation could be either vertical or horizontal.
For vertical hyperbolas, the standard form is given by:

• \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)

Here, the positive term is \( y^2 \), indicating a vertical orientation.

• For horizontal hyperbolas, it's: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)

In these equations, \( a \) and \( b \) are real numbers that determine the size and shape of the hyperbola. The help they provide in calculations is invaluable when identifying other features like vertices and foci.

Vertices of hyperbola

The vertices of a hyperbola are significant points that give insight into its width and general shape. They serve as the closest and furthest points to the center, located along the transverse axis. For a vertical hyperbola, these vertices are vertically aligned.
Using our equation form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the vertices can be easily found at the coordinates:

• \( (0, \pm a) \) for a hyperbola centered at (0,0)

Here, \( a \) is derived from the term \( \frac{y^2}{a^2} \). Calculating \( a \) involves taking the square root of \( a^2 \), identified from the standard equation.
For example, in the equation \( \frac{y^2}{4} - \frac{x^2}{81} = 1 \), \( a^2 = 4 \), thus \( a = 2 \), giving us vertices at \( (0,2) \) and \( (0,-2) \). These values tell how far up and down the graph will reach from the origin.

Foci of hyperbola

The foci or focus points of a hyperbola are crucial as they have a direct influence on its eccentricity and shape. They lie along the same axis as the vertices but are located outside the transverse length of the hyperbola.
The distance \( c \) to each focus from the center is given by:

• \( c = \sqrt{a^2 + b^2} \)

Here, \( a^2 \) is the denominator of the positive term and \( b^2 \) is the other denominator in the standard form. For explicit identification in the equation \( \frac{y^2}{4} - \frac{x^2}{81} = 1 \):

• \( c = \sqrt{4 + 81} = \sqrt{85} \)
• Foci: \( (0, \pm \sqrt{85}) \)

By understanding the placement of the foci, one appreciates the symmetry of the hyperbola and comprehends how it envelopes the conic section.

Equations of asymptotes

Asymptotes in a hyperbola are straight lines that approach the curves of the hyperbola at infinity. Despite never touching the hyperbola, they provide an excellent guide about its end-behavior. For a hyperbola, its shape and opening direction is sketched based upon these asymptotes.
For our vertical hyperbola form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), asymptotes follow the equations:

• \( y = \pm \frac{a}{b}x \)

Substituting values from the equation \( \frac{y^2}{4} - \frac{x^2}{81} = 1 \), gives:

• \( a = 2 \), \( b = 9 \)
• Asymptote equations become \( y = \pm \frac{2}{9}x \)

These lines, cross the center of the hyperbola (origin for our case) defining its steepness and how wide it appears when graphed, offering a scaffold on how the hyperbola opens towards infinity.