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]
