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What is the center of the hyperbola x2−9y2−81=0x^2 - 9y^2 - 81 = 0?\newline(_,_)(\_,\_)

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Q. What is the center of the hyperbola x2−9y2−81=0x^2 - 9y^2 - 81 = 0?\newline(_,_)(\_,\_)
  1. Move constant term: x2−9y2−81=0x^2 - 9y^2 - 81 = 0\newlineMove the constant term to the right side.\newlinex2−9y2=81x^2 - 9y^2 = 81
  2. Convert to standard form: x2−9y2=81x^2 - 9y^2 = 81\newlineConvert the equation into standard form.\newlineDivide both sides of the equation by 8181.\newlinex281−9y281=8181\frac{x^2}{81} - \frac{9y^2}{81} = \frac{81}{81}\newlinex2/81−y2/9=1x^2/81 - y^2/9 = 1
  3. Find center of hyperbola: x281−y29=1\frac{x^2}{81} - \frac{y^2}{9} = 1\newlineFind the center of the hyperbola.\newlineThe standard form of the equation of a hyperbola centered at (h,k)(h, k) is (x−h)2a2−(y−k)2b2=1\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1.\newlineHere, the equation can be written as (x−0)281−(y−0)29=1\frac{(x - 0)^2}{81} - \frac{(y - 0)^2}{9} = 1.\newlineThus, h=0h = 0 and k=0k = 0.\newlineCenter of the hyperbola: (0,0)(0, 0)

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