Question

find the standard form of the equation of the hyperbola shown.
the standard form of the equation is \boxed{}

Explanation:

Step1: Identify center and orientation

Center at (0,0), opens left/right (horizontal hyperbola). Standard form: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Step2: Find a (distance from center to vertex)

Vertices at (6,0) and (-6,0), so $a=6$, $a^2=36$.

Step3: Find asymptote slope to get b

Asymptotes: $y = \pm\frac{1}{3}x$. Slope $\frac{b}{a} = \frac{1}{3}$. Since $a=6$, $b=2$, $b^2=4$.

Step4: Write standard equation

Substitute $a^2=36$ and $b^2=4$ into the form.

Answer:

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