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What is the center of the hyperbola $16x^2 - y^2 = 64$ ? $\newline$ $(\_,\_)$

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Find properties of hyperbolas from equations in general form

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Q. What is the center of the hyperbola

16x^2 - y^2 = 64

?

\newline

(\_,\_)

Write Equation: Write the given equation of the hyperbola. $\newline$ The given equation is $16x^2 - y^2 = 64$ .
Rearrange Equation: Rearrange the equation to resemble the standard form of a hyperbola. $\newline$ To do this, we need to isolate the terms with $x$ and $y$ on one side and move the constant to the other side. We have: $\newline$ $16x^2 - y^2 - 64 = 0$ $\newline$ Adding $64$ to both sides gives us: $\newline$ $16x^2 - y^2 = 64$
Divide Equation: Divide the equation by $64$ to get the standard form of the hyperbola. $\newline$ Dividing each term by $64$ , we get: $\newline$ $\frac{16x^2}{64} - \frac{y^2}{64} = \frac{64}{64}$ $\newline$ Simplifying this, we have: $\newline$ $x^2/4 - y^2/64 = 1$
Identify Center: Identify the center of the hyperbola. $\newline$ The standard form of a hyperbola is $(x-h)^2/a^2 - (y-k)^2/b^2 = 1$ , where $(h, k)$ is the center of the hyperbola. In our equation $x^2/4 - y^2/64 = 1$ , we can see that $h = 0$ and $k = 0$ , since there are no terms to shift the hyperbola left/right or up/down. $\newline$ Therefore, the center of the hyperbola is $(0, 0)$ .

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