To find the equation of the hyperbola, we follow these steps:

Step 1: General Equation and Orientation

The asymptotes $y = \pm \frac{1}{3}x$ suggest the hyperbola has a center at the origin ($h = 0, k = 0$), and it is oriented horizontally since one vertex is given as $(6, 0)$. For a horizontal hyperbola, the equation is:

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

Step 2: Relation Between $a$, $b$, and the Slope of Asymptotes

The slopes of the asymptotes are $\pm \frac{b}{a} = \pm \frac{1}{3}$. Therefore:

$\frac{b}{a} = \frac{1}{3} \implies b = \frac{a}{3}.$

Step 3: Use Vertex to Determine $a$

The distance from the center to the vertex is equal to $a$. Since one vertex is at $(6, 0)$, we have:

$a = 6.$

Step 4: Find $b^2$

Using $b = \frac{a}{3}$, we calculate:

$b = \frac{6}{3} = 2 \implies b^2 = 4.$

Step 5: Write the Equation

Substituting $a^2 = 36$ and $b^2 = 4$ into the equation of the hyperbola, we get:

$\frac{x^2}{36} - \frac{y^2}{4} = 1.$

Final Answer:

The equation of the hyperbola is:

$\boxed{\frac{x^2}{36} - \frac{y^2}{4} = 1.}$

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Would you like a detailed explanation of the derivation, or do you have any questions?

Related Questions:

1. How do you derive the equation of hyperbolas with vertical orientation?
2. What are the key features (foci, asymptotes, etc.) of this hyperbola?
3. How do the slopes of the asymptotes affect the orientation of a hyperbola?
4. How can we find the foci of this hyperbola?
5. How do you generalize the equation of a hyperbola with a center not at the origin?

Tip:

The distance between the foci of a hyperbola can be calculated using $2c$, where $c = \sqrt{a^2 + b^2}$. This is useful for fully characterizing the hyperbola.