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Equation:$$16x^2-9y^2+32x+36y-164=0$$

I know that it's a hyperbola by reducing it to this: $$\frac{(x+1)^2}{9}-\frac{(y-2)^2}{16}=1$$

My question is that this is different from what i have studied in school to be hyperbola's equation:

$$\frac{(x)^2}{a^2}-\frac{(y)^2}{b^2}=1$$

So why is it so?

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    $\begingroup$ The equation you studied in school is for a hyperbola centered at the origin. The equation you found is a hyperbola centered at $(x,y)=(-1,2)$. $\endgroup$
    – Servaes
    Apr 27, 2017 at 11:11

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Uhh, i don't know your basics about conics. But this simply indicates about the shift of center from origin $(0,0)$ to $(-1,2)$.

Theory: $$\frac{(x-a)^2}{A^2}-\frac{(x-b)^2}{B^2}=1$$ means that the hyperbola is now centered at $(a,b)$.

Hope It solve's your problem.

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It's not any different, just that center is shifted to (-1,2).

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If the function can be expressed $f(x-a,y-b)$, $f(x,y)$ move from $(0,0)$ to $(a,b)$.

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