Problem 14

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. $$ 9 y^{2}-4 x^{2}=1 $$

Step-by-Step Solution

Verified

Answer

Vertices: (0, 1/3) and (0, -1/3); Foci: (0, √13/6) and (0, -√13/6); Asymptotes: y = ±(2/3)x.

1Step 1: Identify the Standard Form

The standard form for a hyperbola oriented vertically is $ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $. The given equation is $ 9y^2 - 4x^2 = 1 $. We can compare this with the standard form of a hyperbola to recognize it is already in standard form.

2Step 2: Write in Standard Form

Divide each term by 1 to express the equation as $ \frac{9y^2}{1} - \frac{4x^2}{1} = 1 $. Rewrite it as $ \frac{y^2}{\frac{1}{9}} - \frac{x^2}{\frac{1}{4}} = 1 $. Now it fits the form $ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $, with $ a^2 = \frac{1}{9} $ and $ b^2 = \frac{1}{4} $.

3Step 3: Calculate the Vertices

The vertices occur at $ y = \pm a $ for a hyperbola of this form. Here, $ a = \sqrt{\frac{1}{9}} = \frac{1}{3} $. Therefore, the vertices are at $ (0, \frac{1}{3}) $ and $ (0, -\frac{1}{3}) $.

4Step 4: Calculate the Foci

The foci are found at $ y = \pm c $ where $ c = \sqrt{a^2 + b^2} $. First, calculate $ c^2 = \frac{1}{9} + \frac{1}{4} = \frac{4}{36} + \frac{9}{36} = \frac{13}{36} $. Thus, $ c = \sqrt{\frac{13}{36}} = \frac{\sqrt{13}}{6} $. The foci are at $ (0, \frac{\sqrt{13}}{6}) $ and $ (0, -\frac{\sqrt{13}}{6}) $.

5Step 5: Determine the Asymptotes

For a hyperbola with vertical transverse axis, the asymptotes are given by $ y = \pm \frac{a}{b} x $. Here, $ a = \frac{1}{3} $ and $ b = \frac{1}{2} $. Therefore, the equations of the asymptotes are $ y = \pm \frac{\frac{1}{3}}{\frac{1}{2}} x = \pm \frac{2}{3} x $.

Key Concepts

Standard FormVerticesFociAsymptotes

Standard Form

When working with hyperbolas, we often want to express them in 'standard form'. This provides a clear framework for understanding their structure and locating key features like vertices, foci, and asymptotes. For a hyperbola oriented vertically, the standard form is given by:

$ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $

Our example to consider is the hyperbola given by the equation:\[ 9y^2 - 4x^2 = 1 \]This aligns with the standard form, as the positive term is $ y^2 $ and the negative term is $ x^2 $. By rewriting it distinctly, i.e.,

$ \frac{y^2}{(1/9)} - \frac{x^2}{(1/4)} = 1 $

we identify $ a^2 = \frac{1}{9} $ and $ b^2 = \frac{1}{4} $. This set-up allows us to easily identify and calculate other essential characteristics of the hyperbola.

Vertices

The vertices of a hyperbola are the points where it intersects its transverse axis, the axis that determines the orientation of the hyperbola. In the standard form

$ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $

we find the vertices along the y-axis at $ y = \pm a $.

For the given hyperbola, calculate $ a $ as

$ a = \sqrt{\frac{1}{9}} = \frac{1}{3} $

Thus, the vertices are located at the points

$ (0, \frac{1}{3}) $
$ (0, -\frac{1}{3}) $

"Understanding the location of the vertices is crucial," as they reflect the extremes of the hyperbola's width.

Foci

The foci of a hyperbola are points located on the interior of the branches that aid in the geometric definition of the hyperbola. They lie along the transverse axis, in this case the y-axis. The distance from the center to each focus, $ c $, is calculated with the formula:

$ c = \sqrt{a^2 + b^2} $

For our hyperbola,

Calculate $ c^2 = \frac{1}{9} + \frac{1}{4} = \frac{13}{36} $
Then $ c = \sqrt{\frac{13}{36}} = \frac{\sqrt{13}}{6} $

This places the foci at

$ (0, \frac{\sqrt{13}}{6}) $
$ (0, -\frac{\sqrt{13}}{6}) $

Locating the foci is essential for accurately sketching the hyperbola.

Asymptotes

Asymptotes are lines that a hyperbola approaches but never meets. They provide a boundary that the curves get infinitely close to without crossing. For a vertically oriented hyperbola, the equations of asymptotes can be written as: