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What is the center of the hyperbola x2−4y2−64=0x^2 - 4y^2 - 64 = 0?\newline(_,_)(\_,\_)

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Q. What is the center of the hyperbola x2−4y2−64=0x^2 - 4y^2 - 64 = 0?\newline(_,_)(\_,\_)
  1. Move constant term: Move the constant term to the right side of the equation.\newlinex2−4y2−64=0x^2 - 4y^2 - 64 = 0 becomes x2−4y2=64x^2 - 4y^2 = 64.
  2. Convert to standard form: Convert the equation into standard form by dividing both sides by 6464. \newlinex264−4y264=6464\frac{x^2}{64} - \frac{4y^2}{64} = \frac{64}{64} simplifies to x264−y216=1\frac{x^2}{64} - \frac{y^2}{16} = 1.
  3. Identify center: Identify the center of the hyperbola.\newlineThe equation x264−y216=1\frac{x^2}{64} - \frac{y^2}{16} = 1 can be written as (x−0)2/64−(y−0)2/16=1\left(x - 0\right)^2/64 - \left(y - 0\right)^2/16 = 1.\newlineHere h=0h = 0 and k=0k = 0, so the center of the hyperbola is at the point (0,0)(0, 0).

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