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What is the center of the hyperbola $x^2 - 4y^2 = 64$ ? $\newline$ $(\_,\_$ )

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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

x^2 - 4y^2 = 64

?

\newline

(\_,\_

)

Write Equation: Write the given equation of the hyperbola. $\newline$ The given equation is $x^2 - 4y^2 = 64$ .
Rearrange for Center: Rearrange the equation to identify the center. $\newline$ To find the center, we need to express the equation in the standard form of a hyperbola. The standard form for a horizontal hyperbola is $(x-h)^2/a^2 - (y-k)^2/b^2 = 1$ , where $(h, k)$ is the center of the hyperbola. For a vertical hyperbola, the terms are switched.
Divide by $64$ : Divide the equation by $64$ to get the standard form. $\newline$ Divide both sides of the equation by $64$ to get $\frac{x^2}{64} - \frac{4y^2}{64} = 1$ . $\newline$ Simplify to get $\frac{x^2}{64} - \frac{y^2}{16} = 1$ .
Identify Center: Identify the center of the hyperbola. $\newline$ From the standard form $\frac{x^2}{64} - \frac{y^2}{16} = 1$ , we can see that the equation can be written as $\left(x-0\right)^2/64 - \left(y-0\right)^2/16 = 1$ . Therefore, the center $(h, k)$ of the hyperbola is $(0, 0)$ .

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