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Mrac [35]
3 years ago
6

Find the vertices and foci of the hyperbola with equation quantity x minus five squared divided by eighty one minus the quantity

of y minus one squared divided by one hundred and forty four = 1.
Mathematics
1 answer:
Papessa [141]3 years ago
3 0
If you notice, the positive fraction, is the one with the "x" variable in it, that simply means, the hyperbola is opening over the x-axis, so is a horizontal one, therefore

\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}{81}-\cfrac{(y-1)^2}{144}=1\implies \cfrac{(x-5)^2}{9^2}-\cfrac{(y-1)^2}{12^2}=1~~
\begin{cases}
a=9\\
b=12
\end{cases}
\\\\\\
c=\sqrt{9^2+12^2}\implies c=15\\\\
-------------------------------\\\\
vertices~~(5\pm 9,1)\implies 
\begin{cases}
(14,1)\\
(-4,1)
\end{cases}~~foci~~(5\pm 15,1)\implies 
\begin{cases}
(20,1)\\
(-10,1)
\end{cases}
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