To solve for the equations of the directrices of the given hyperbola, follow these steps:
1. Identify and understand the given information about the hyperbola:
- The center of the hyperbola is at the origin [tex]\((0, 0)\)[/tex].
- One vertex is at [tex]\((-6, 0)\)[/tex].
- One focus is at [tex]\((10, 0)\)[/tex].
2. Determine the values for [tex]\(a\)[/tex] and [tex]\(c\)[/tex]:
- The vertices are located at [tex]\((-a, 0)\)[/tex] and [tex]\((a, 0)\)[/tex]. Given one vertex at [tex]\((-6, 0)\)[/tex], we have [tex]\(a = 6\)[/tex]. Thus, the vertices are [tex]\( (6, 0) \)[/tex] and [tex]\(( -6, 0) \)[/tex].
- The foci are located at [tex]\((-c, 0)\)[/tex] and [tex]\((c, 0)\)[/tex]. Given one focus at [tex]\((10, 0)\)[/tex], we have [tex]\(c = 10\)[/tex]. Thus, the foci are [tex]\((10, 0)\)[/tex] and [tex]\((-10, 0)\)[/tex].
3. Find [tex]\(b^2\)[/tex] using the relationship [tex]\(c^2 = a^2 + b^2\)[/tex]:
- We know [tex]\(a = 6\)[/tex] and [tex]\(c = 10\)[/tex].
- Using the relationship: [tex]\[ c^2 = a^2 + b^2 \][/tex]
[tex]\[ 10^2 = 6^2 + b^2 \][/tex]
[tex]\[ 100 = 36 + b^2 \][/tex]
[tex]\[ b^2 = 64 \][/tex]
[tex]\[ b = \sqrt{64} = 8 \][/tex]
4. Use the formula for the directrices of a hyperbola:
- The directrices for a hyperbola centered at the origin with horizontal transverse axis are given by: [tex]\[ x = \pm \frac{a^2}{c} \][/tex]
- Substitute the known values [tex]\(a = 6\)[/tex] and [tex]\(c = 10\)[/tex]:
[tex]\[ \text{Directrices: } x = \pm \frac{6^2}{10} \][/tex]
[tex]\[ x = \pm \frac{36}{10} \][/tex]
[tex]\[ x = \pm 3.6 \][/tex]
5. State the final result:
- The equations of the directrices for the given hyperbola are:
[tex]\[ x = \pm \frac{3}{5}\][/tex]
Thus, the directrices of the hyperbola are [tex]\( x = \pm \frac{3}{5} \)[/tex].
