For $$ y^{2}-\frac{(x-6)^{2}}{20}=1 $$, (a) Identify the center. (b) Identify the vertices. (c) Identify the foci. (d) Write equations for the asymptotes. (e) Graph the hyperbola. Give exact answers. Express numbers in simplest form using integers, fractions, or radicals. Part 1 of 5 (a) The center is $$ (6,0) $$. Part 2 of 5 (b) The vertices are $$ (6, \boxed{1}) $$ and $$ (6, \boxed{-1}) $$. Part 3 of 5 (c) The foci are $$ (6, \sqrt{21}) $$ and $$ (6,-\sqrt{21}) $$. Part: $$ 3 / 5 $$ Part 4 of 5 (d) The asymptotes are $$ y=\square $$ and $$ y= $$ $$ \square $$ .

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UpStudy AI · Accepted Answer

The hyperbola is written in the standard form  
$$
\frac{(y-0)^2}{1^2} - \frac{(x-6)^2}{(2\sqrt{5})^2} = 1,
$$  
with center $$(6,0)$$, $$a=1$$, and $$b=2\sqrt{5}$$.

For a vertical hyperbola the asymptotes are given by  
$$
y - k = \pm \frac{a}{b}(x - h).
$$

Substituting $$h=6$$, $$k=0$$, $$a=1$$, and $$b=2\sqrt{5}$$ gives  
$$
y = \pm \frac{1}{2\sqrt{5}} (x-6).
$$

Thus, the asymptote equations are  
$$
y=\frac{1}{2\sqrt{5}}(x-6) \quad \text{and} \quad y=-\frac{1}{2\sqrt{5}}(x-6).
$$
The asymptotes are $$ y = \frac{1}{2\sqrt{5}}(x-6) $$ and $$ y = -\frac{1}{2\sqrt{5}}(x-6) $$.

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