Question 860695
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Hi,
re TY: Foci {{{sqrt(a^2+b^2)}}}units units up and down from center
there is NO minus to consider (You may be thinking of the ellipse format)
hyperbola centered at (1, -4), with one focus at (7, -4) and eccentricity e=2.
Center, Vertices and foci along y = -4 
Yes!
{{{(x-1)^2/a^2-(y+4)^2/b^2 = 1}}}
2 = c/a and Yes,  c = 6  and a = 3
{{{(x-1)^2/3^2-(y+4)^2/b^2 = 1}}}
 sqrt(9+ 27) = 6, b^2 = 27   |Foci {{{sqrt(a^2+b^2)}}}units units up and down from center
{{{(x-1)^2/9-(y+4)^2/27 = 1}}}