Asked by Anonymous
what is the foci of the graph x^2/36-y^2/16=1
Answers
Answered by
drwls
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
(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
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