QuestionFind the center, vertices, foci, and the equations of the asymptotes of the hyperbola. (If an answer does not exist, enter DNE.) $$\frac{(x+4)^{2}}{144}-\frac{(y-2)^{2}}{25}=1$$

Studdy Solution
Derive the equations of the asymptotes:
For a hyperbola that opens horizontally, the equations of the asymptotes are:
$y = k \pm \frac{b}{a}(x-h)$
Substitute the known values:
$y = 2 \pm \frac{5}{12}(x + 4)$
Simplify the equations:
1. $y = 2 + \frac{5}{12}(x + 4)$
2. $y = 2 - \frac{5}{12}(x + 4)$
These are the equations of the asymptotes.
The center is $(-4, 2)$, the vertices are $(-16, 2)$ and $(8, 2)$, the foci are $(-17, 2)$ and $(9, 2)$, and the equations of the asymptotes are $y = 2 + \frac{5}{12}(x + 4)$ and $y = 2 - \frac{5}{12}(x + 4)$.