Given that the hyperbola is centered at the origin and has vertical vertices and foci, the relevant information aligns with the second column in your table.
The vertex is at [tex]\((0,36)\)[/tex], so we have [tex]\(a = 36\)[/tex].
The focus is at [tex]\((0,39)\)[/tex], so we have [tex]\(c = 39\)[/tex].
For a hyperbola with vertical transverse axis, the equations of the directrices are given by:
[tex]\[ y = \pm \frac{a^2}{c} \][/tex]
Here's the detailed step-by-step solution to determine the equations of the directrices:
1. Calculate [tex]\(a^2\)[/tex]:
[tex]\[a^2 = 36^2 = 1296\][/tex]
2. Calculate [tex]\( \frac{a^2}{c} \)[/tex]:
[tex]\[ \frac{a^2}{c} = \frac{1296}{39} \approx 33.23076923076923\][/tex]
3. Write the equations of the directrices:
[tex]\[y = \pm \frac{a^2}{c}\][/tex]
Which gives us the two equations:
[tex]\[ y = 33.23076923076923 \][/tex]
[tex]\[ y = -33.23076923076923 \][/tex]
Therefore, the equations of the directrices are [tex]\( y = 33.23076923076923 \)[/tex] and [tex]\( y = -33.23076923076923 \)[/tex].
