Question

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.

Answer 1

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 √( 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=√(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}`