Question

Find the vertices and foci of the hyperbola with equation quantity x minus 5 squared divided by 144 minus the quantity of y minus 4 squared divided by 81 = 1.
A. Vertices: (14, 4), (-4, 4); Foci: (-4, 4), (14, 4)
B. Vertices: (17, 4), (-7, 4); Foci: (-10, 4), (20, 4)
C. Vertices: (4, 17), (4, -7); Foci: (4, -10), (4, 20)
D. Vertices: (4, 14), (4, -4); Foci: (4, -4), (4, 14)

Answer 1

$\bf \textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases}\\\\ -------------------------------$

$\bf \cfrac{(x-5)^2}{144}-\cfrac{(y-4)^2}{81}=1\implies \cfrac{(x-5)^2}{12^2}-\cfrac{(y-4)^2}{9^2}=1 \\\\\\ \begin{cases} h=5\\ k=4\\ a=12\\ b=9 \end{cases}\implies c=\sqrt{144+81}\implies c=\sqrt{225}\implies c=15 \\\\\\ \stackrel{center}{(5,4)}\qquad \qquad \stackrel{\textit{because is a horizontal traverse hyperbola}}{\stackrel{foci}{\stackrel{(5\pm 15,4)}{(20,4),(-10,4)}}\qquad \qquad \stackrel{vertices}{\stackrel{(5\pm 12,4)}{(17,4),(-7,4)}}}$