Question

(Y-2)^2/16-(x+1)^2/144+1

Answer 1

`\textit{hyperbolas, vertical traverse axis } \\\\ \cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h, k\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad √( a ^2 + b ^2) \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(y-2)^2}{16}-\cfrac{(x+1)^2}{144}=1\implies \cfrac{(y-2)^2}{4^2}-\cfrac{( ~~ x-(-1) ~~ )^2}{12^2}=1`

now, since the positive fraction is the one with the "y" in it, that means the traverse axis is running over the y-axis in short the hyperbola is opening up and down.

From the same equation above, we can see the center is at (-1 , 2) and the conjugate axis has a half which is "b" or namely b = 12, since that's half of the conjugate axis, then the conjugate axis must be 2b or 24.