For $$ y^{2}-\frac{(x+5)^{2}}{18}=1 $$, (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 $$ -5,0 $$ ). Part 2 of 5 (b) The vertices are $$ (-5,1) $$ and $$ (-5,-1) $$. Part 3 of 5 (c) The foci are $$ (\sqrt{-5}, \sqrt{19}) $$ and $$ (\boxed{-5},-\sqrt{19}) $$. 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 given in standard form as

$$
y^{2} - \frac{(x+5)^2}{18} = 1.
$$

For a vertical hyperbola in the form

$$
\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1,
$$

the center is $$(h,k)$$, the vertices are $$(h,k \pm a)$$, the foci are $$(h,k \pm c)$$ with $$c^2 = a^2+b^2$$, and the asymptotes are given by

$$
y-k = \pm \frac{a}{b} (x-h).
$$

For the given hyperbola:

- The center is $$(-5, 0)$$.
- The value for $$a^2$$ is 1, so $$a = 1$$.
- The value for $$b^2$$ is 18, so $$b = \sqrt{18} = 3\sqrt{2}$$.
- To find $$c$$, compute

$$
c^2 = a^2 + b^2 = 1 + 18 = 19, \quad c = \sqrt{19}.
$$

Thus, the foci are located at

$$
(-5, 0 \pm \sqrt{19}) \quad \Rightarrow \quad (-5, \sqrt{19}) \text{ and } (-5, -\sqrt{19}).
$$

The equations for the asymptotes, using the center $$(-5,0)$$, are

$$
y = 0 \pm \frac{a}{b}(x+5).
$$

Since $$\frac{a}{b} = \frac{1}{3\sqrt{2}}$$, the asymptotes are

$$
y = \frac{1}{3\sqrt{2}}(x+5) \quad \text{and} \quad y = -\frac{1}{3\sqrt{2}}(x+5).
$$

It is often preferable to rationalize the coefficient:

$$
\frac{1}{3\sqrt{2}} = \frac{\sqrt{2}}{6}.
$$

Thus, the equations for the asymptotes can also be written as

$$
y = \frac{\sqrt{2}}{6}(x+5) \quad \text{and} \quad y = -\frac{\sqrt{2}}{6}(x+5).
$$

So the answers for part (d) are:

- $$ y = \frac{\sqrt{2}}{6}(x+5) $$
- $$ y = -\frac{\sqrt{2}}{6}(x+5) $$
(a) Center: $$(-5, 0)$$ (b) Vertices: $$(-5, 1)$$ and $$(-5, -1)$$ (c) Foci: $$(-5, \sqrt{19})$$ and $$(-5, -\sqrt{19})$$ (d) Asymptotes: - $$ y = \frac{\sqrt{2}}{6}(x + 5) $$ - $$ y = -\frac{\sqrt{2}}{6}(x + 5) $$ (e) Graph of the hyperbola is not provided here.

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