Respuesta :
[tex] \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}\\\\
------------------------------- [/tex]
[tex] \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)}}} [/tex]
The vertices: (17, 4), (-7, 4); Foci: (-10, 4), (20, 4) of the hyperbola.
The correct option is B.
Given
The equation represents the hyperbola;
[tex]\rm \dfrac{(x-5)^2}{144}-\dfrac{(y-4)^2}{81}=1[/tex]
The standard form of hyperbola;
The standard equation of hyperbola can be represented as;
[tex]\rm \dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1[/tex]
Where h and k are the centers of the hyperbola.
Converting the equation into standard form hyperbola;
[tex]\rm \dfrac{(x-5)^2}{12^2}-\dfrac{(y-4)^2}{9^2}=1[/tex]
Then,
The value of c is;
[tex]\rm c=\sqrt{a^2+b^2}\\ \\c=\sqrt{9^2+12^2} \\\\ c=\sqrt{81+144} \\\\ c=\sqrt{225} \\\\c=15[/tex]
Therefore,
The vertices and foci of the hyperbola are;
[tex]\rm Vertices =(h\pm a, \ k) = (5\pm 12, \ 4)= (17, 4)(-7, 4)\\\\Foci= (h\pm c, k)= (5 \pm15, 4)=(20,4)(-10, 4)[/tex]
Hence, the vertices: (17, 4), (-7, 4); Foci: (-10, 4), (20, 4) of the hyperbola.
To know more about Hyperbola click the link given below.
https://brainly.com/question/9479638
