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]
