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A hyperbola centered at the origin has a vertex at [tex]\((-6,0)\)[/tex] and a focus at [tex]\((10,0)\)[/tex].

\begin{tabular}{|l|l|}
\hline
Vertices: [tex]\((-a, 0), (a, 0)\)[/tex] & Vertices: [tex]\((0, -a), (0, a)\)[/tex] \\
Foci: [tex]\((-c, 0), (c, 0)\)[/tex] & Foci: [tex]\((0, -c), (0, c)\)[/tex] \\
Asymptotes: [tex]\(y= \pm \frac{b}{a} x\)[/tex] & Asymptotes: [tex]\(y= \pm \frac{a}{b} x\)[/tex] \\
Directrices: [tex]\(x= \pm \frac{a^2}{c}\)[/tex] & Directrices: [tex]\(y= \pm \frac{a^2}{c}\)[/tex] \\
Standard Equation: [tex]\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)[/tex] & Standard Equation: [tex]\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)[/tex] \\
\hline
\end{tabular}

Which are the equations of the directrices?

[tex]\[ x= \pm \frac{a^2}{c} \][/tex]



Answer :

GinnyAnswer
Let's determine the equations of the directrices given for the hyperbola centered at the origin, with a vertex at [tex]\((-6, 0)\)[/tex] and a focus at [tex]\((10, 0)\)[/tex].

### Step-by-Step Solution

1. Identifying Parameters:
- Vertices: [tex]\((-a, 0)\)[/tex] and [tex]\((a, 0)\)[/tex]. Given the vertex at [tex]\((-6, 0)\)[/tex], we have [tex]\(a = 6\)[/tex].
- Foci: [tex]\((-c, 0)\)[/tex] and [tex]\((c, 0)\)[/tex]. Given the focus at [tex]\((10, 0)\)[/tex], we have [tex]\(c = 10\)[/tex].

2. Calculating [tex]\(a^2\)[/tex] and [tex]\(c^2\)[/tex]:
- [tex]\(a^2 = 6^2 = 36\)[/tex]
- [tex]\(c^2 = 10^2 = 100\)[/tex]

3. Finding Directrix Equation:
- The directrix for a hyperbola centered at the origin with a horizontal transverse axis is given by [tex]\(x = \pm \frac{a^2}{c}\)[/tex].

4. Plugging in Values:
- [tex]\( \frac{a^2}{c} = \frac{36}{10} = 3.6 \)[/tex]

Therefore, the equations of the directrices are:
[tex]\[ x = \pm 3.6 \][/tex]

Since your provided equation [tex]\(x= \pm \frac{3}{5}\)[/tex] does not match this correct result of [tex]\(x = \pm 3.6\)[/tex], it is evident that the equations of the directrices for this hyperbola are indeed:
[tex]\[ x = \pm 3.6 \][/tex]

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