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.
Accepted Solution
A:
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
[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}\\\\
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