what is the foci of the graph x^2/36-y^2/16=1

Answer 1

That is the equation of a hyperbola.

(x/6)^2 - (y/4)^2 = 1

= (x/a)^2 - (y/b)^2

where a = 6 and b = 4.

The foci are at (-c, 0) and (c,0), where

c = sqrt (a^2 + b^2) = sqrt 52 = 7.21