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

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

Which are the equations of the asymptotes?

A. \(y= \pm \frac{9}{40} x\)
B. [tex]\(y= \pm \frac{40}{1} x\)[/tex]



Answer :

GinnyAnswer
To determine the equations of the asymptotes for the given hyperbola, let's follow a systematic approach.

Given:
- The vertex is at \((0, -40)\) which implies \(a = 40\).
- The focus is at \((0, 41)\) which implies \(c = 41\).

For hyperbolas centered at the origin with a vertical transverse axis and the standard form equation \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the relationship between \(a\), \(b\), and \(c\) is given by:
[tex]\[ c^2 = a^2 + b^2 \][/tex]

Given that \(a = 40\) and \(c = 41\), we can calculate \(b\):

[tex]\[ c^2 = a^2 + b^2 \][/tex]
[tex]\[ 41^2 = 40^2 + b^2 \][/tex]
[tex]\[ 1681 = 1600 + b^2 \][/tex]
[tex]\[ b^2 = 1681 - 1600 \][/tex]
[tex]\[ b^2 = 81 \][/tex]
[tex]\[ b = \sqrt{81} \][/tex]
[tex]\[ b = 9 \][/tex]

The slopes of the asymptotes for hyperbolas with a vertical transverse axis are given by \(\pm \frac{a}{b}\):

[tex]\[ \frac{a}{b} = \frac{40}{9} \][/tex]

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

So, the correct answer is:
[tex]\[ y = \pm \frac{40}{9} x \][/tex]

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