find the center of the hyperbola defined by this equation: (y-6)^2/9-(x+8)^2/64=1

Question

amistre64 · Answer

centers are always attached to your xy parts

amistre64 · Answer

(x-Cx) + (y-Cy) = n

such that (Cx,Cy) is the center point

anonymous · Answer

so is that the answer

amistre64 · Answer

im pretty sure it is ...

amistre64 · Answer

what attached to your x part?

hero · Answer

That's not the answer. It's a general form

anonymous · Answer

i need the answer

anonymous · Answer

plz

amistre64 · Answer

then tell me what is attached to the x part so we can come to the particular answer ...

anonymous · Answer

thats all that is given in the question

amistre64 · Answer

yes, so lets read what is given ...

amistre64 · Answer

(x+ ?? )

what do we see attached to the x part?

anonymous · Answer

i dont no!! whatever is in that question above is all that is given i dont no what x is that is just what is asked thats what i dont understand

amistre64 · Answer

we are not trying to determine what x IS ; but rather what is sitting next to it.

anonymous · Answer

it wants to no the center of the hyperbola, the answer should be coordinates

amistre64 · Answer

yes, and the coordinates are given in the equation itself; we just have to know where they are sitting in order to simply pluck them from the equation.

anonymous · Answer

well idk then cuz thats all that is attatched tothis question there is nothing else

amistre64 · Answer

the equation given is this:

 (y-6)^2/9-(x+8)^2/64=1

lets strip away all the useless noise to get at what we want

 (y-6)  (x+8)

now we are left with the parts that we need to get an answer

amistre64 · Answer

what is attached to the x?  and what is attached to the y?

anonymous · Answer

omg idont know

hero · Answer

amistre, by he way, you didn't show him the correct general form.

anonymous · Answer

i just told you that all was in the question i dont know anything else

amistre64 · Answer

yeah, im sure thats where they are having problems at ...

amistre64 · Answer

you should review your material that shows you how to construct these conic equations ... other than that I dont know what else I can actually do that would be helpful.

hero · Answer

General Form for hyperbola: 
$$\frac{(y-k)^2}{a^2} - \frac{(x-h)}{b^2} = 1$$ where (h,k) is the center

In your case, you have: 

$$\frac{(y-6)^2}{3^2} - \frac{(x-(-8))^2}{8^2} = 1$$ where (-8,6) is the center