What is the coordinates of the center of #y^2/25-(x-6)^2/144=1#?

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Answer 1 · 2017-06-13T16:56:23

The center is #(6,0)#

The standard form for a hyperbola of this type is:

#(y-k)^2/a^2-(x-h)^2/b^2=1" [1]"#

where #(h,k)# is the center.

We can observe from the given equation,

#y^2/25-(x-6)^2/144=1" [2]"#

that #k = 0# and #h = 6#

Therefore the center is #(6,0)#