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A hyperbola centered at the origin has a vertex at [tex]\((0,36)\)[/tex] and a focus at [tex]\((0,39)\)[/tex].

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

Which are the equations of the directrices?



Answer :

GinnyAnswer
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].

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