Problem 13 - College Algebra

Question

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

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

Step 1: Identify the Equation Type

The given equation is \( \frac{y^2}{4} - \frac{x^2}{81} = 1 \). This equation represents a hyperbola in the standard form because it follows the format \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) for a hyperbola with a vertical transverse axis.

Step 2: Identify Transverse and Conjugate Axes

For the given equation \( \frac{y^2}{4} - \frac{x^2}{81} = 1 \), the term \( \frac{y^2}{4} \) tells us this hyperbola opens vertically. Here, \( a^2 = 4 \) and \( b^2 = 81 \), so \( a = 2 \) and \( b = 9 \).

Step 3: Determine the Vertices

The vertices of a vertically-oriented hyperbola are \( (0, \pm a) \). For this hyperbola, \( a = 2 \), so the vertices are at \( (0, 2) \) and \( (0, -2) \).

Step 4: Find the Foci

The distance from the center to each focus is given by \( c \), where \( c^2 = a^2 + b^2 \). Here, \( c^2 = 4 + 81 = 85 \), thus \( c = \sqrt{85} \). The foci are \( (0, \pm \sqrt{85}) \).

Step 5: Write the Equations of the Asymptotes

For a vertical hyperbola, the equations of the asymptotes are \( y = \pm \frac{a}{b}x \). Substituting \( a = 2 \) and \( b = 9 \), the equations of the asymptotes are \( y = \pm \frac{2}{9}x \).

Key Concepts

Hyperbola Equation

The equation of a hyperbola comes in a specific standard form. In the exercise provided, the equation is \(\frac{y^2}{4} - \frac{x^2}{81} = 1\), which indicates a hyperbola. Hyperbolas have two main forms based on their orientation:

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

In this case, since \( y^2 \) is first and positive, this is a vertically oriented hyperbola. The terms under the squares help determine the hyperbola's shape and orientation. Here, \(a^2 = 4\) and \(b^2 = 81\), leading to \(a = 2\) and \(b = 9\). A positive term preceding \(y^2\) confirms the vertical orientation of the (\(y\)-axis) hyperbola.

Vertices of a Hyperbola

The vertices of a hyperbola are critical points that define its shape in space. For a vertical hyperbola, the vertices are based on the value of \(a\), which is associated with the \(y\)-term.

• The vertices are located at \((0, \pm a)\) when the center is at the origin.

For the exercise equation \(\frac{y^2}{4} - \frac{x^2}{81} = 1\), \(a = 2\). Therefore, the vertices are positioned at \((0, 2)\) and \((0, -2)\), indicating they lie on the vertical axis, which the hyperbola opens along.

Foci of a Hyperbola

The foci of a hyperbola are points that signify extreme distances from the center to help define the hyperbola's shape.

• For vertical hyperbolas, foci are located at \((0, \pm c)\), where \(c\) is calculated using the relation \(c^2 = a^2 + b^2\).

In this problem, \(a^2 = 4\) and \(b^2 = 81\), so \(c^2 = 4 + 81 = 85\), giving \(c = \sqrt{85}\). Thus, the foci points are at \((0, \sqrt{85})\) and \((0, -\sqrt{85})\). These points ensure the hyperbola forms its characteristic curves.

Asymptotes of a Hyperbola

The asymptotes of a hyperbola are diagonal lines that the hyperbola approaches but never intersects. These lines help in sketching the general direction and spread of the hyperbola.

• For vertically oriented hyperbolas, the asymptotes are represented by the equations \(y = \pm \frac{a}{b}x\).

In this instance, substituting \(a = 2\) and \(b = 9\) into the formula gives the asymptote equations: \(y = \frac{2}{9}x\) and \(y = -\frac{2}{9}x\). These equations describe the slope of the lines confirming that the branches of the hyperbola extend towards these lines but never touch them.