For $$ y^{2}-\frac{(x-5)^{2}}{12}=1 $$, $$ \begin{array}{ll} ext { (a) Identify the center. } & ext { (b) Identify the vertices. } \ \begin{array}{ll} ext { (c) Identify the foci. } & ext { (d) Write equations for the asymptotes. } \ ext { (e) Graph the hyperbola. }\end{array} \ ext { Give exact answers. Express numbers in simplest form using integers, fractions, or radicals. }\end{array} $$

Question

For $$ y^{2}-\frac{(x-5)^{2}}{12}=1 $$, $$ \begin{array}{ll}	ext { (a) Identify the center. } & 	ext { (b) Identify the vertices. } \ \begin{array}{ll}	ext { (c) Identify the foci. } & 	ext { (d) Write equations for the asymptotes. } \ 	ext { (e) Graph the hyperbola. }\end{array} \ 	ext { Give exact answers. Express numbers in simplest form using integers, fractions, or radicals. }\end{array} $$

UpStudy AI · Accepted Answer

**(a) Identify the center**

The given hyperbola is 
$$
y^{2}-\frac{(x-5)^{2}}{12}=1.
$$
We can rewrite it in the standard form for a vertical transverse axis:
$$
\frac{(y-0)^2}{1^2}-\frac{(x-5)^2}{(\sqrt{12})^2}=1.
$$
Thus, the center is at 
$$
(5,0).
$$

---

**(b) Identify the vertices**

For a hyperbola in the form 
$$
\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1,
$$
the vertices lie along the vertical line $$x=h$$ at a distance $$a$$ from the center. Here, $$a^2=1$$ so $$a=1$$. Therefore, the vertices are:
$$
(5,0+1)=(5,1) \quad \text{and} \quad (5,0-1)=(5,-1).
$$

---

**(c) Identify the foci**

For a vertical hyperbola, the foci are given by 
$$
(h, k\pm c),
$$
where 
$$
c=\sqrt{a^2+b^2}.
$$
We have $$a^2=1$$ and $$b^2=12$$, so:
$$
c=\sqrt{1+12}=\sqrt{13}.
$$
Thus, the foci are:
$$
(5,0+\sqrt{13})=(5,\sqrt{13}) \quad \text{and} \quad (5,0-\sqrt{13})=(5,-\sqrt{13}).
$$

---

**(d) Write equations for the asymptotes**

The asymptotes of a vertical hyperbola in the form 
$$
\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1
$$
are given by:
$$
y-k=\pm \frac{a}{b}(x-h).
$$
Here, $$a=1$$ and $$b=\sqrt{12}=2\sqrt{3}$$, and the center is $$(5,0)$$. Thus, the equations become:
$$
y=\pm \frac{1}{2\sqrt{3}}(x-5).
$$
Alternatively, by rationalizing the denominator:
$$
\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6},
$$
the asymptotes can be written as:
$$
y=\pm \frac{\sqrt{3}}{6}(x-5).
$$

---

**(e) Graph the hyperbola**

To graph the hyperbola, follow these steps:

1. **Plot the Center:**  
   Mark the point $$(5,0)$$.

2. **Plot the Vertices:**  
   Mark the points $$(5,1)$$ and $$(5,-1)$$.

3. **Plot the Foci:**  
   Mark the points $$(5,\sqrt{13})$$ and $$(5,-\sqrt{13})$$.  
   (Note that $$\sqrt{13}$$ is approximately $$3.61$$.)

4. **Draw the Asymptotes:**  
   Draw straight lines through the center with equations  
   $$
   y=\frac{1}{2\sqrt{3}}(x-5) \quad \text{and} \quad y=-\frac{1}{2\sqrt{3}}(x-5).
   $$  
   These lines guide the opening of the hyperbola.

5. **Sketch the Hyperbola:**  
   Draw two branches of the hyperbola that open upwards and downwards. The branches approach, but never cross, the asymptotes as $$x$$ moves away from 5.

The resulting graph shows a hyperbola with a vertical transverse axis, centered at $$(5,0)$$ with its vertices at $$(5,1)$$ and $$(5,-1)$$, and with asymptotes given by the lines above.
**(a) Center:** $$(5, 0)$$ **(b) Vertices:** $$(5, 1)$$ and $$(5, -1)$$ **(c) Foci:** $$(5, \sqrt{13})$$ and $$(5, -\sqrt{13})$$ **(d) Asymptotes:** $$ y = \frac{1}{2\sqrt{3}}(x - 5) \quad \text{and} \quad y = -\frac{1}{2\sqrt{3}}(x - 5) $$ **(e) Graph:** A vertical hyperbola centered at $$(5, 0)$$ with vertices at $$(5, 1)$$ and $$(5, -1)$$, foci at $$(5, \sqrt{13})$$ and $$(5, -\sqrt{13})$$, and asymptotes $$y = \frac{1}{2\sqrt{3}}(x - 5)$$ and $$y = -\frac{1}{2\sqrt{3}}(x - 5)$$.

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