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What is the center of the hyperbola x29y2=81x^2 - 9y^2 = 81?\newline(_,_)(\_,\_)

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Q. What is the center of the hyperbola x29y2=81x^2 - 9y^2 = 81?\newline(_,_)(\_,\_)
  1. Move constant to right: x29y2=81x^2 - 9y^2 = 81\newlineMove the constant term to the right side to prepare the equation for standard form.\newlinex29y281+81=81x^2 - 9y^2 - 81 + 81 = 81\newlinex29y2=81x^2 - 9y^2 = 81
  2. Convert to standard form: x29y2=81x^2 - 9y^2 = 81\newlineConvert the equation into standard form by dividing both sides by 8181.\newlinex2819y281=8181\frac{x^2}{81} - \frac{9y^2}{81} = \frac{81}{81}\newlinex2/81y2/9=1x^2/81 - y^2/9 = 1
  3. Identify center: x281y29=1\frac{x^2}{81} - \frac{y^2}{9} = 1\newlineIdentify the center of the hyperbola.\newlineThe equation can be written as (x0)2/81(y0)2/9=1(x - 0)^2/81 - (y - 0)^2/9 = 1.\newlineHere h=0h = 0, k=0k = 0.\newlineCenter of the hyperbola: (0,0)(0, 0)

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