To find the equations of the asymptotes for the given hyperbola centered at the origin with a vertex at [tex]\((9, 0)\)[/tex] and a focus at [tex]\((-15, 0)\)[/tex], we follow these steps:
1. Identify Values of [tex]\(a\)[/tex] and [tex]\(c\)[/tex] - The given vertex is at [tex]\((9, 0)\)[/tex]. Therefore, [tex]\(a = 9\)[/tex] since the vertex is [tex]\( \pm a \)[/tex] units away from the center. - The given focus is at [tex]\((-15, 0)\)[/tex]. Since foci are at [tex]\( \pm c \)[/tex], the distance [tex]\(c = 15\)[/tex]. Note that the value will be taken as positive since distance is non-negative.
2. Calculate [tex]\(b\)[/tex] - For a hyperbola with the standard equation [tex]\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)[/tex], the relationship between [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] is given by: [tex]\[
c^2 = a^2 + b^2
\][/tex] - Substituting [tex]\(a = 9\)[/tex] and [tex]\(c = 15\)[/tex] into the equation: [tex]\[
15^2 = 9^2 + b^2
\][/tex] [tex]\[
225 = 81 + b^2
\][/tex] [tex]\[
b^2 = 225 - 81
\][/tex] [tex]\[
b^2 = 144
\][/tex] [tex]\[
b = \sqrt{144} = 12
\][/tex]
3. Find the Equations of the Asymptotes - For hyperbolas with horizontal transverse axes centered at the origin, the equations of the asymptotes are given by: [tex]\[
y = \pm \frac{b}{a} x
\][/tex] - Substituting [tex]\(a = 9\)[/tex] and [tex]\(b = 12\)[/tex]: [tex]\[
y = \pm \frac{12}{9} x
\][/tex] [tex]\[
y = \pm \frac{4}{3} x
\][/tex]
So, the equations of the asymptotes are: [tex]\[ y = \frac{4}{3} x \][/tex] [tex]\[ y = -\frac{4}{3} x \][/tex]
However, the given prompt suggests [tex]\( y = \pm \frac{3}{5} x \)[/tex], which does not match the correct derived answer. The correct equations of the asymptotes based on our calculations should indeed be: [tex]\[ y = \pm \frac{4}{3} x \][/tex]
