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A hyperbola centered at the origin has a vertex at [tex]\((0, -40)\)[/tex] and a focus at [tex]\((0, 41)\)[/tex].

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

Which are the equations of the asymptotes?

A. [tex]\(y = \pm \frac{9}{40} x\)[/tex]

B. [tex]\(y = \pm \frac{40}{41} x\)[/tex]

Sagot :

GinnyAnswer
Let's solve the problem step-by-step to find the equations of the asymptotes for the hyperbola.

Given:
- The hyperbola is centered at the origin.
- The vertex of the hyperbola is at [tex]\((0, -40)\)[/tex].
- The focus of the hyperbola is at [tex]\((0, 41)\)[/tex].

From this information, the hyperbola has its transverse axis along the [tex]\(y\)[/tex]-axis, which means we will use the second set of equations from the table provided.

The general form of the equation for a hyperbola centered at the origin with a vertical transverse axis is:
[tex]\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \][/tex]

Where:
- [tex]\(a\)[/tex] is the distance from the center to a vertex.
- [tex]\(c\)[/tex] is the distance from the center to a focus.
- [tex]\(b\)[/tex] is given by the relationship [tex]\(b^2 = c^2 - a^2\)[/tex].

Here, we have:
- [tex]\(a = 40\)[/tex] (since the vertex is at [tex]\((0, -40)\)[/tex]).
- [tex]\(c = 41\)[/tex] (since the focus is at [tex]\((0, 41)\)[/tex]).

We first find [tex]\(b\)[/tex] using the equation [tex]\(b^2 = c^2 - a^2\)[/tex]:
[tex]\[ b^2 = 41^2 - 40^2 \][/tex]
[tex]\[ b^2 = 1681 - 1600 \][/tex]
[tex]\[ b^2 = 81 \][/tex]
[tex]\[ b = \sqrt{81} = 9 \][/tex]

Now, the equations of the asymptotes for a hyperbola with a vertical transverse axis are:
[tex]\[ y = \pm \frac{a}{b}x \][/tex]

So, substituting the values we have:
[tex]\[ y = \pm \frac{40}{9}x \][/tex]

Therefore, the equations of the asymptotes are:
[tex]\[ y = \pm \frac{40}{9} x \][/tex]

Which simplify to:
[tex]\[ y = \pm 4.444444444444445 x \][/tex]

So out of the given options, this solves to neither of the suggested forms as they might have typographical errors. The proper asymptote equations derived from our accurate calculation would indeed be:
[tex]\[ y = \pm 4.444444444444445 x. \][/tex]
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