What are the vertices of the hyperbola given by the equation (y-4)^2/121 - (x+9)^2/141 = 1?

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What are the vertices of the hyperbola given by the equation (y-4)^2/121 - (x+9)^2/141 = 1?

Mathematics
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first you need the center so you know that one ?
No, I really don't understand this section which is why I was asking for help.
center of \[\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\]is \((h,k)\)

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Other answers:

in your case it is \((-9,4)\)
because the \(y\) part comes first (before the minus sign) this hyperbola looks like this (roughly) |dw:1353293873616:dw|
to find the vertices, go \(a\) units up up and down from the center. that is why you needed the center first in your case \(a^2=121\) and so \(a=11\)
since the center is \((-9,4)\) if you go up 11 units you are at \((-9,15)\) and if you go down 11 units you are at \((-9,-7)\)
hope that is clear, and also that you see it is not that hard once you know what you are doing
I'm still extremely lost, as you can tell math isn't my strong point. I appreciate the help however.

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